Poj 1273 Drainage Ditches(最大流 Edmonds-Karp )

题目链接:poj1273 Drainage Ditches

呜呜,今天自学网络流,看了EK算法,学的晕晕的,留个简单模板题来作纪念、、、

 #include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<set>
#include<algorithm>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std; const int N = ;
const int inf = 0x3f3f3f3f;
int n, m;
int g[N][N];//i到j的容量
int pre[N];//i的前驱节点
bool vis[N];
int flow[N][N];//从i到j的流量
int a[N];//从源点到i节点的最小残留量 int Edmonds_Karp(int st, int ed){
int u, v;
int f = ;//最大流
queue<int>q;//bfs找增广路
while(){
CLR(vis, ); CLR(a, );//每找一次记得初始化
q.push(st);
vis[st] = ;
a[st] = inf;
while(!q.empty()){
u = q.front(); q.pop();
for(v = ; v <= n; ++v){
if(!vis[v] && g[u][v] > flow[u][v]){
q.push(v);
pre[v] = u;
vis[v] = ;
a[v] = min(a[u], g[u][v] - flow[u][v]);
}
}
}
if(a[ed] == ) break;//当前已经是最大流
f += a[ed];
for(u = ed; u != st; u = pre[u]){
flow[pre[u]][u] += a[ed];
flow[u][pre[u]] -= a[ed];
} }
return f;
}
int main(){
int i, j, k, a, b, c;
while(scanf("%d %d", &m, &n)==){
CLR(g, ); CLR(flow, );
for(i = ; i < m; ++i){
scanf("%d %d %d", &a, &b, &c);
g[a][b] += c;//可能有重边
}
int max_flow = Edmonds_Karp(, n);
printf("%d\n", max_flow);
}
return ;
}
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