I am trying to solve the following graph problem:
We are given a general unweighted and undirected graph and k (k < |V| ) vertices that are
already known beforehand. The vertices are deleted sequentially. After
each deletion, how many connected components are there?
I thought of using tarjan's algorithm at each step to check if the current vertex to be deleted is a cut vertex so that when the deletion is performed, we can simply add the number of neighbours to the number of connected components. The complexity of this algorithm is O(V(V+E)).
I was told that there is a O(V+E) algorithm to perform this task. But I cannot figure it out. Research on Google also does not reveal much. Could anyone please advise me?
We can use the fact that the vertices are known beforehand.
Let's solve a "reverse" problem: given a graph and a list vertices that are ADDED to it sequentially, compute the number of connected components in the graph after each addition structure.
The solution is pretty straightforward: we can maintain a disjoint set union structure and add all edges incident to the vertex to the graph (it's easy to keep the number of components in this structure: initially, it is equal to the number of vertices and is decreased by one when a union actually happens).
The original problem is reduced to the "reverse" one in the following way:
Let's add all edges that are not incident to any of the deleted vertices to the disjoint set union.
Now we can reverse the list of deleted vertices and add them one by one as described above.
After that, we need to reverse the resulting list that contains the number of components.
Note: this solution is not actually O(V + E), its O(V + E * alpha(V)), where alpha(x) is the inverse Ackermann's function. It is very close to linear for all practical purposes.
here is my implementation of algorithm in c++ using disjoint set:
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
typedef pair<int, int> pii;
const int M=2e5+137;
class DisjointSet {
public:
int connected_comp;
int parent[100000];
void makeSet(int n){
for (int i=1;i<n+1; ++i)
parent[i] = i;
connected_comp = n;
}
int Find(int l) {
if (parent[l] == l)
return l;
return Find(parent[l]);
}
void Union(int m, int n) {
int x = Find(m);
int y = Find(n);
if(x==y) return;
if(x<y){
parent[y] = x;
connected_comp--;
}
else{
parent[x] = y;
connected_comp--;
}
}
};
set<pii> not_delete;
vector<pii> to_add;
int main(){
int node, edge;
cout<<"enter number of nodes and edges"<<"\n";
cin>>node>>edge;
DisjointSet dis;
dis.makeSet(node);
cout<<"enter two nodes to add edges"<<"\n";
for(int i=0;i<edge;i++){
int u,v;
cin>>u>>v;
if(u>v){
not_delete.insert({u,v});
}
else{
not_delete.insert({v,u});
}
}
int deletions;
cout<<"enter number of deletions"<<"\n";
cin>>deletions;
cout<<"enter two node to delete edge between them"<<"\n";
for(int i=0;i<deletions;i++){
int u,v;
cin>>u>>v;
if(u>v){
not_delete.erase({u,v});// edges that never delete from graph
to_add.pb({u,v}); // edges that gonna delete from graph
}
else{
not_delete.erase({v,u});
to_add.pb({v,u});
}
}
vector<int> res;
// first adding edges that never delete from graph
for(pii x: not_delete){
dis.Union(x.first, x.second);
}
res.pb(dis.connected_comp);
// then adding edges that will be deleted from graph backwards
reverse(to_add.begin(), to_add.end());
for(pii x: to_add){
dis.Union(x.first, x.second);
res.pb(dis.connected_comp);
}
cout<<"connected components after each deletion:"<<"\n";
for (auto it = ++res.rbegin(); it != res.rend(); ++it)
cout << *it << "\n";
return 0;
}
A classic algorithm question in 2D version is typically described as
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
For example, Given the input
[0,1,0,2,1,0,1,3,2,1,2,1]
the return value would be
6
The algorithm that I used to solve the above 2D problem is
int trapWaterVolume2D(vector<int> A) {
int n = A.size();
vector<int> leftmost(n, 0), rightmost(n, 0);
//left exclusive scan, O(n), the highest bar to the left each point
int leftMaxSoFar = 0;
for (int i = 0; i < n; i++){
leftmost[i] = leftMaxSoFar;
if (A[i] > leftMaxSoFar) leftMaxSoFar = A[i];
}
//right exclusive scan, O(n), the highest bar to the right each point
int rightMaxSoFar = 0;
for (int i = n - 1; i >= 0; i--){
rightmost[i] = rightMaxSoFar;
if (A[i] > rightMaxSoFar) rightMaxSoFar = A[i];
}
// Summation, O(n)
int vol = 0;
for (int i = 0; i < n; i++){
vol += max(0, min(leftmost[i], rightmost[i]) - A[i]);
}
return vol;
}
My Question is how to make the above algorithm extensible to the 3D version of the problem, to compute the maximum of water trapped in real-world 3D terrain. i.e. To implement
int trapWaterVolume3D(vector<vector<int> > A);
Sample graph:
We know the elevation at each (x, y) point and the goal is to compute the maximum volume of water that can be trapped in the shape. Any thoughts and references are welcome.
For each point on the terrain consider all paths from that point to the border of the terrain. The level of water would be the minimum of the maximum heights of the points of those paths. To find it we need to perform a slightly modified Dijkstra's algorithm, filling the water level matrix starting from the border.
For every point on the border set the water level to the point height
For every point not on the border set the water level to infinity
Put every point on the border into the set of active points
While the set of active points is not empty:
Select the active point P with minimum level
Remove P from the set of active points
For every point Q adjacent to P:
Level(Q) = max(Height(Q), min(Level(Q), Level(P)))
If Level(Q) was changed:
Add Q to the set of active points
user3290797's "slightly modified Dijkstra algorithm" is closer to Prim's algorithm than Dijkstra's. In minimum spanning tree terms, we prepare a graph with one vertex per tile, one vertex for the outside, and edges with weights equal to the maximum height of their two adjoining tiles (the outside has height "minus infinity").
Given a path in this graph to the outside vertex, the maximum weight of an edge in the path is the height that the water has to reach in order to escape along that path. The relevant property of a minimum spanning tree is that, for every pair of vertices, the maximum weight of an edge in the path in the spanning tree is the minimum possible among all paths between those vertices. The minimum spanning tree thus describes the most economical escape paths for water, and the water heights can be extracted in linear time with one traversal.
As a bonus, since the graph is planar, there's a linear-time algorithm for computing the minimum spanning tree, consisting of alternating Boruvka passes and simplifications. This improves on the O(n log n) running time of Prim.
This problem can be solved using the Priority-Flood algorithm. It's been discovered and published a number of times over the past few decades (and again by other people answering this question), though the specific variant you're looking for is not, to my knowledge, in the literature.
You can find a review paper of the algorithm and its variants here. Since that paper was published an even faster variant has been discovered (link), as well as methods to perform this calculation on datasets of trillions of cells (link). A method for selectively breaching low/narrow divides is discussed here. Contact me if you'd like copies of any of these papers.
I have a repository here with many of the above variants; additional implementations can be found here.
A simple script to calculate volume using the RichDEM library is as follows:
#include "richdem/common/version.hpp"
#include "richdem/common/router.hpp"
#include "richdem/depressions/Lindsay2016.hpp"
#include "richdem/common/Array2D.hpp"
/**
#brief Calculates the volume of depressions in a DEM
#author Richard Barnes (rbarnes#umn.edu)
Priority-Flood starts on the edges of the DEM and then works its way inwards
using a priority queue to determine the lowest cell which has a path to the
edge. The neighbours of this cell are added to the priority queue if they
are higher. If they are lower, then they are members of a depression and the
elevation of the flooding minus the elevation of the DEM times the cell area
is the flooded volume of the cell. The cell is flooded, total volume
tracked, and the neighbors are then added to a "depressions" queue which is
used to flood depressions. Cells which are higher than a depression being
filled are added to the priority queue. In this way, depressions are filled
without incurring the expense of the priority queue.
#param[in,out] &elevations A grid of cell elevations
#pre
1. **elevations** contains the elevations of every cell or a value _NoData_
for cells not part of the DEM. Note that the _NoData_ value is assumed to
be a negative number less than any actual data value.
#return
Returns the total volume of the flooded depressions.
#correctness
The correctness of this command is determined by inspection. (TODO)
*/
template <class elev_t>
double improved_priority_flood_volume(const Array2D<elev_t> &elevations){
GridCellZ_pq<elev_t> open;
std::queue<GridCellZ<elev_t> > pit;
uint64_t processed_cells = 0;
uint64_t pitc = 0;
ProgressBar progress;
std::cerr<<"\nPriority-Flood (Improved) Volume"<<std::endl;
std::cerr<<"\nC Barnes, R., Lehman, C., Mulla, D., 2014. Priority-flood: An optimal depression-filling and watershed-labeling algorithm for digital elevation models. Computers & Geosciences 62, 117–127. doi:10.1016/j.cageo.2013.04.024"<<std::endl;
std::cerr<<"p Setting up boolean flood array matrix..."<<std::endl;
//Used to keep track of which cells have already been considered
Array2D<int8_t> closed(elevations.width(),elevations.height(),false);
std::cerr<<"The priority queue will require approximately "
<<(elevations.width()*2+elevations.height()*2)*((long)sizeof(GridCellZ<elev_t>))/1024/1024
<<"MB of RAM."
<<std::endl;
std::cerr<<"p Adding cells to the priority queue..."<<std::endl;
//Add all cells on the edge of the DEM to the priority queue
for(int x=0;x<elevations.width();x++){
open.emplace(x,0,elevations(x,0) );
open.emplace(x,elevations.height()-1,elevations(x,elevations.height()-1) );
closed(x,0)=true;
closed(x,elevations.height()-1)=true;
}
for(int y=1;y<elevations.height()-1;y++){
open.emplace(0,y,elevations(0,y) );
open.emplace(elevations.width()-1,y,elevations(elevations.width()-1,y) );
closed(0,y)=true;
closed(elevations.width()-1,y)=true;
}
double volume = 0;
std::cerr<<"p Performing the improved Priority-Flood..."<<std::endl;
progress.start( elevations.size() );
while(open.size()>0 || pit.size()>0){
GridCellZ<elev_t> c;
if(pit.size()>0){
c=pit.front();
pit.pop();
} else {
c=open.top();
open.pop();
}
processed_cells++;
for(int n=1;n<=8;n++){
int nx=c.x+dx[n];
int ny=c.y+dy[n];
if(!elevations.inGrid(nx,ny)) continue;
if(closed(nx,ny))
continue;
closed(nx,ny)=true;
if(elevations(nx,ny)<=c.z){
if(elevations(nx,ny)<c.z){
++pitc;
volume += (c.z-elevations(nx,ny))*std::abs(elevations.getCellArea());
}
pit.emplace(nx,ny,c.z);
} else
open.emplace(nx,ny,elevations(nx,ny));
}
progress.update(processed_cells);
}
std::cerr<<"t Succeeded in "<<std::fixed<<std::setprecision(1)<<progress.stop()<<" s"<<std::endl;
std::cerr<<"m Cells processed = "<<processed_cells<<std::endl;
std::cerr<<"m Cells in pits = " <<pitc <<std::endl;
return volume;
}
template<class T>
int PerformAlgorithm(std::string analysis, Array2D<T> elevations){
elevations.loadData();
std::cout<<"Volume: "<<improved_priority_flood_volume(elevations)<<std::endl;
return 0;
}
int main(int argc, char **argv){
std::string analysis = PrintRichdemHeader(argc,argv);
if(argc!=2){
std::cerr<<argv[0]<<" <Input>"<<std::endl;
return -1;
}
return PerformAlgorithm(argv[1],analysis);
}
It should be straight-forward to adapt this to whatever 2d array format you are using
In pseudocode, the following is equivalent to the foregoing:
Let PQ be a priority-queue which always pops the cell of lowest elevation
Let Closed be a boolean array initially set to False
Let Volume = 0
Add all the border cells to PQ.
For each border cell, set the cell's entry in Closed to True.
While PQ is not empty:
Select the top cell from PQ, call it C.
Pop the top cell from PQ.
For each neighbor N of C:
If Closed(N):
Continue
If Elevation(N)<Elevation(C):
Volume += (Elevation(C)-Elevation(N))*Area
Add N to PQ, but with Elevation(C)
Else:
Add N to PQ with Elevation(N)
Set Closed(N)=True
This problem is very close to the construction of the morphological watershed of a grayscale image.
One approach is as follows (flooding process):
sort all pixels by increasing elevation.
work incrementally, by increasing elevations, assigning labels to the pixels per catchment basin.
for a new elevation level, you need to label a new set of pixels:
Some have no labeled
neighbor, they form a local minimum configuration and begin a new catchment basin.
Some have only neighbors with the same label, they can be labeled similarly (they extend a catchment basin).
Some have neighbors with different labels. They do not belong to a specific catchment basin and they define the watershed lines.
You will need to enhance the standard watershed algorithm to be able to compute the volume of water. You can do that by determining the maximum water level in each basin and deduce the ground height on every pixel. The water level in a basin is given by the elevation of the lowest watershed pixel around it.
You can act every time you discover a watershed pixel: if a neighboring basin has not been assigned a level yet, that basin can stand the current level without leaking.
In order to accomplish tapping water problem in 3D i.e., to calculate the maximum volume of trapped rain water you can do something like this:
#include<bits/stdc++.h>
using namespace std;
#define MAX 10
int new2d[MAX][MAX];
int dp[MAX][MAX],visited[MAX][MAX];
int dx[] = {1,0,-1,0};
int dy[] = {0,-1,0,1};
int boundedBy(int i,int j,int k,int in11,int in22)
{
if(i<0 || j<0 || i>=in11 || j>=in22)
return 0;
if(new2d[i][j]>k)
return new2d[i][j];
if(visited[i][j]) return INT_MAX;
visited[i][j] = 1;
int r = INT_MAX;
for(int dir = 0 ; dir<4 ; dir++)
{
int nx = i + dx[dir];
int ny = j + dy[dir];
r = min(r,boundedBy(nx,ny,k,in11,in22));
}
return r;
}
void mark(int i,int j,int k,int in1,int in2)
{
if(i<0 || j<0 || i>=in1 || j>=in2)
return;
if(new2d[i][j]>=k)
return;
if(visited[i][j]) return ;
visited[i][j] = 1;
for(int dir = 0;dir<4;dir++)
{
int nx = i + dx[dir];
int ny = j + dy[dir];
mark(nx,ny,k,in1,in2);
}
dp[i][j] = max(dp[i][j],k);
}
struct node
{
int i,j,key;
node(int x,int y,int k)
{
i = x;
j = y;
key = k;
}
};
bool compare(node a,node b)
{
return a.key>b.key;
}
vector<node> store;
int getData(int input1, int input2, int input3[])
{
int row=input1;
int col=input2;
int temp=0;
int count=0;
for(int i=0;i<row;i++)
{
for(int j=0;j<col;j++)
{
if(count==(col*row))
break;
new2d[i][j]=input3[count];
count++;
}
}
store.clear();
for(int i = 0;i<input1;i++)
{
for(int j = 0;j<input2;j++)
{
store.push_back(node(i,j,new2d[i][j]));
}
}
memset(dp,0,sizeof(dp));
sort(store.begin(),store.end(),compare);
for(int i = 0;i<store.size();i++)
{
memset(visited,0,sizeof(visited));
int aux = boundedBy(store[i].i,store[i].j,store[i].key,input1,input2);
if(aux>store[i].key)
{
memset(visited,0,sizeof(visited));
mark(store[i].i,store[i].j,aux,input1,input2);
}
}
long long result =0 ;
for(int i = 0;i<input1;i++)
{
for(int j = 0;j<input2;j++)
{
result = result + max(0,dp[i][j]-new2d[i][j]);
}
}
return result;
}
int main()
{
cin.sync_with_stdio(false);
cout.sync_with_stdio(false);
int n,m;
cin>>n>>m;
int inp3[n*m];
store.clear();
for(int j = 0;j<n*m;j++)
{
cin>>inp3[j];
}
int k = getData(n,m,inp3);
cout<<k;
return 0;
}
class Solution(object):
def trapRainWater(self, heightMap):
"""
:type heightMap: List[List[int]]
:rtype: int
"""
m = len(heightMap)
if m == 0:
return 0
n = len(heightMap[0])
if n == 0:
return 0
visited = [[False for i in range(n)] for j in range(m)]
from Queue import PriorityQueue
q = PriorityQueue()
for i in range(m):
visited[i][0] = True
q.put([heightMap[i][0],i,0])
visited[i][n-1] = True
q.put([heightMap[i][n-1],i,n-1])
for j in range(1, n-1):
visited[0][j] = True
q.put([heightMap[0][j],0,j])
visited[m-1][j] = True
q.put([heightMap[m-1][j],m-1,j])
S = 0
while not q.empty():
cell = q.get()
for (i, j) in [(1,0), (-1,0), (0,1), (0,-1)]:
x = cell[1] + i
y = cell[2] + j
if x in range(m) and y in range(n) and not visited[x][y]:
S += max(0, cell[0] - heightMap[x][y]) # how much water at the cell
q.put([max(heightMap[x][y],cell[0]),x,y])
visited[x][y] = True
return S
Here is the simple code for the same-
#include<iostream>
using namespace std;
int main()
{
int n,count=0,a[100];
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
}
for(int i=1;i<n-1;i++)
{
///computing left most largest and Right most largest element of array;
int leftmax=0;
int rightmax=0;
///left most largest
for(int j=i-1;j>=1;j--)
{
if(a[j]>leftmax)
{
leftmax=a[j];
}
}
///rightmost largest
for(int k=i+1;k<=n-1;k++)
{
if(a[k]>rightmax)
{
rightmax=a[k];
}
}
///computing hight of the water contained-
int x=(min(rightmax,leftmax)-a[i]);
if(x>0)
{
count=count+x;
}
}
cout<<count;
return 0;
}
I have a class Graph with two lists types namely nodes and edges
I have a function
List<int> GetNodesInRange(Graph graph, int Range)
when I get these parameters I need an algorithm that will go through the graph and return the list of nodes only as deep (the level) as the range.
The algorithm should be able to accommodate large number of nodes and large ranges.
Atop this, should I use a similar function
List<int> GetNodesInRange(Graph graph, int Range, int selected)
I want to be able to search outwards from it, to the number of nodes outwards (range) specified.
alt text http://www.freeimagehosting.net/uploads/b110ccba58.png
So in the first function, should I pass the nodes and require a range of say 2, I expect the results to return the nodes shown in the blue box.
The other function, if I pass the nodes as in the graph with a range of 1 and it starts at node 5, I want it to return the list of nodes that satisfy this criteria (placed in the orange box)
What you need seems to be simply a depth-limited breadth-first search or depth-first search, with an option of ignoring edge directionality.
Here's a recursive definition that may help you:
I'm the only one of range 1 from myself.
I know who my immediate neighbors are.
If N > 1, then those of range N from myself are
The union of all that is of range N-1 from my neighbors
It should be a recursive function, that finds neighbours of the selected, then finds neighbours of each neighbour until range is 0. DFS search something like that:
List<int> GetNodesInRange(Graph graph, int Range, int selected){
var result = new List<int>();
result.Add( selected );
if (Range > 0){
foreach ( int neighbour in GetNeighbours( graph, selected ) ){
result.AddRange( GetNodesInRange(graph, Range-1, neighbour) );
}
}
return result;
}
You should also check for cycles, if they are possible. This code is for tree structure.
// get all the nodes that are within Range distance of the root node of graph
Set<int> GetNodesInRange(Graph graph, int Range)
{
Set<int> out = new Set<int>();
GetNodesInRange(graph.root, int Range, out);
return out;
}
// get all the nodes that are within Range successor distance of node
// accepted nodes are placed in out
void GetNodesInRange(Node node, int Range, Set<int> out)
{
boolean alreadyVisited = out.add(node.value);
if (alreadyVisited) return;
if (Range == 0) return;
// for each successor node
{
GetNodesInRange(successor, Range-1, out);
}
}
// get all the nodes that are within Range distance of selected node in graph
Set<int> GetNodesInRange(Graph graph, int Range, int selected)
{
Set<int> out = new Set<int>();
GetNodesInRange(graph, Range, selected, out);
return out;
}
// get all the nodes that are successors of node and within Range distance
// of selected node
// accepted nodes are placed in out
// returns distance to selected node
int GetNodesInRange(Node node, int Range, int selected, Set<int> out)
{
if (node.value == selected)
{
GetNodesInRange(node, Range-1, out);
return 1;
}
else
{
int shortestDistance = Range + 1;
// for each successor node
{
int distance = GetNodesInRange(successor, Range, selected, out);
if (distance < shortestDistance) shortestDistance = distance;
}
if (shortestDistance <= Range)
{
out.add(node.value);
}
return shortestDistance + 1;
}
}
I modified your requirements somewhat to return a Set rather than a List.
The GetNodesInRange(Graph, int, int) method will not handle graphs that contain cycles. This can be overcome by maintaining a collection of nodes that have already been visited. The GetNodesInRange(Graph, int) method makes use of the fact that the out set is a collection of visited nodes to overcome cycles.
Note: This has not been tested in any way.