题意：有m个点，n条管道，问从1到m最大能够同时通过的水量是多少？
题解：最大流模板题。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 0x3ffffff;
const int maxn = 205;
int gra[maxn][maxn];
int flow[maxn];
int path[maxn];
int n,m;
int bfs(int s, int e)
{
queue<int>q;
int t;
memset(path,-1,sizeof(path));
memset(flow,0,sizeof(flow));
path[s] = 0;
flow[s] = INF;
q.push(s);
while(!q.empty()){
t = q.front();
q.pop();
if(t == e) break;
for(int i = 1; i <= m; i ++)
{
if(i != s && gra[t][i] && path[i] == -1)
{
flow[i] = flow[t] < gra[t][i] ? flow[t] : gra[t][i];
path[i] = t;
q.push(i);
}
}
}
if(path[e] == -1)
return -1;
else return flow[e];
}
int EK(int s, int e){
int maxflow = 0;
int pre, now, step;
while((step = bfs(s,e))!= -1){
maxflow += step;
now = e;
while(now != s){
pre = path[now];
gra[pre][now] -= step;
gra[now][pre] += step;
now = pre;
}
}
return maxflow;
}
int main()
{
int u,v,c;
while(~scanf("%d%d",&n,&m)){
memset(gra,0,sizeof(gra));
for(int i = 1; i <= n; i ++)
{
scanf("%d %d %d", &u,&v,&c);
gra[u][v] += c;
}
int ans = EK(1,m);
printf("%d
",ans);
}
return 0;
}