Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 7250 Accepted Submission(s): 3364
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)
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
题意：n个节点，m条路径。当中给出每条路径的容量。
求1到n的最大流
<span style="font-size:24px;">#include<stdio.h>
#include<algorithm>
#include<queue>
#define INF 1000
#define min(x,y) (x<y?x:y)
using namespace std;
int cap[20][20],flow[20][20],a[1000],p[100];
int n,m,f,s,t;
void Edmonds_Karp()
{
queue<int> q;
memset(flow,0,sizeof(flow));
f=0;
s=1;
t=n;
while(1)
{
memset(a,0,sizeof(a));
a[s]=INF;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int v=1;v<=n;v++)
{
if(!a[v]&&cap[u][v]>flow[u][v])
{
p[v]=u;
q.push(v);
a[v]=min(a[u],cap[u][v]-flow[u][v]);
}
}
}
if(a[t]==0) break;
for(int u=t;u!=s;u=p[u])
{
flow[p[u]][u]+=a[t];
flow[u][p[u]]-=a[t];
}
f+=a[t];
}
}
int main()
{
int t,k=1;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
memset(cap,0,sizeof(cap));
for(int i=0;i<m;i++)
{
int x,y,c;
scanf("%d %d %d",&x,&y,&c);
cap[x][y]+=c;
}
Edmonds_Karp();
printf("Case %d: %d
",k++,f);
}
return 0;
}</span>