#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 1e9
using namespace std;
queue<int>q;
int mapp[200][200];
int flow[200][200];
int a[200];
int p[200];
int n,m;
int min(int a,int b)
{
if(a>b)return b;
return a;
}
int max_flow(int be,int en)
{
int f,u,v;
memset(flow,0,sizeof(flow));
f = 0;
while(1)
{
memset(a,0,sizeof(a));
a[be] = N;
q.push(be);
while(!q.empty())
{
u = q.front();
q.pop();
for(v=1;v<=m;v++)
{
if(!a[v]&&mapp[u][v]>flow[u][v])
{
p[v] = u;
q.push(v);
a[v] = min(a[u],mapp[u][v]-flow[u][v]);
}
}
}
if(a[en]==0)break;
for(v=en;v!=be;v=p[v])
{
flow[p[v]][v]+=a[en];
flow[v][p[v]]-=a[en];
}
f+=a[en];
}
return f;
}
int main()
{
int T;
int t = 0;
int i;
int a,b,c;
int flag=0;
scanf("%d",&T);
while(T--)
{
flag++;
scanf("%d%d",&m,&n);
memset(mapp,0,sizeof(mapp));
for(i=0;i<n;i++)
{
scanf("%d%d%d",&a,&b,&c);
mapp[a][b] += c;
}
t = max_flow(1,m);
printf("Case %d: %d\n",flag,t);
}
return 0;
}
F - Flow Problem
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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