Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 8203 Accepted Submission(s): 3817
Problem Description
Network
flow is a well-known difficult problem for ACMers. Given a graph, your
task is to find out the maximum flow for the weighted directed graph.
Input
The first line of input contains an integer T, denoting the number of test cases.
For
each test case, the first line contains two integers N and M, denoting
the number of vertexes and edges in the graph. (2 <= N <= 15, 0
<= M <= 1000)
Next M lines, each line contains three integers
X, Y and C, there is an edge from X to Y and the capacity of it is C. (1
<= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
Sample Output
Case 1: 1
Case 2: 2
Author
HyperHexagon
Source
HyperHexagon‘s Summer Gift (Original tasks)
代码:
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<queue> 5 using namespace std; 6 const int inf=0x3f3f3f3f; 7 int mat[30][30]; 8 int dist[31]; 9 int n,m; 10 int min(int a,int b) 11 { 12 return a<b?a:b; 13 } 14 bool bfs(int st,int to){ 15 memset(dist,-1,sizeof(dist)); 16 queue<int>q; 17 q.push(st); 18 dist[st]=0; 19 int t; 20 while(!q.empty()){ 21 t=q.front(); 22 q.pop(); 23 for(int i=1;i<=n;i++){ 24 if(dist[i]<0&&mat[t][i]>0){ 25 dist[i]=dist[t]+1; 26 if(i==to)return 1; 27 q.push(i); 28 } 29 } 30 } 31 return 0; 32 } 33 int dfs(int st,int to,int flow) 34 { 35 int tem; 36 if(st==to||flow==0) return flow; 37 for(int i=1;i<=n;i++){ 38 if((dist[i]==dist[st]+1)&&mat[st][i]>0&&(tem=dfs(i,to,min(mat[st][i],flow)))) 39 { 40 mat[st][i]-=tem; 41 mat[i][st]+=tem; 42 return tem; 43 } 44 } 45 return 0; 46 } 47 int Dinic(int st,int en) 48 { 49 int ans=0; 50 while(bfs(st,en)) 51 ans+=dfs(st,en,inf); 52 return ans; 53 } 54 int main(){ 55 int cas,i,a,b,c; 56 scanf("%d",&cas); 57 for(i=1;i<=cas;i++){ 58 scanf("%d%d",&n,&m); 59 memset(mat,0,sizeof(mat)); 60 while(m--){ 61 scanf("%d%d%d",&a,&b,&c); 62 mat[a][b]+=c; 63 } 64 printf("Case %d: %d\n",i,Dinic(1,n)); 65 } 66 return 0; 67 }