Poj 1273 Drainage Ditches(最大流 Edmonds-Karp )

Posted 2020-08-09 GraceSkyer

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、、、

 1 #include<cstdio>
 2 #include<vector>
 3 #include<queue>
 4 #include<cstring>
 5 #include<set>
 6 #include<algorithm>
 7 #define CLR(a,b) memset((a),(b),sizeof((a)))
 8 using namespace std;
 9 
10 const int N = 201;
11 const int inf = 0x3f3f3f3f;
12 int n, m;
13 int g[N][N];//i到j的容量
14 int pre[N];//i的前驱节点
15 bool vis[N];
16 int flow[N][N];//从i到j的流量
17 int a[N];//从源点到i节点的最小残留量
18 
19 int Edmonds_Karp(int st, int ed){
20     int u, v;
21     int f = 0;//最大流
22     queue<int>q;//bfs找增广路
23     while(1){
24         CLR(vis, 0); CLR(a, 0);//每找一次记得初始化
25         q.push(st);
26         vis[st] = 1;
27         a[st] = inf;
28         while(!q.empty()){
29             u = q.front(); q.pop();
30             for(v = 1; v <= n; ++v){
31                 if(!vis[v] && g[u][v] > flow[u][v]){
32                     q.push(v);
33                     pre[v] = u;
34                     vis[v] = 1;
35                     a[v] = min(a[u], g[u][v] - flow[u][v]);
36                 }
37             }
38         }
39         if(a[ed] == 0) break;//当前已经是最大流
40         f += a[ed];
41         for(u = ed; u != st; u = pre[u]){
42             flow[pre[u]][u] += a[ed];
43             flow[u][pre[u]] -= a[ed];
44         }
45 
46     }
47     return f;
48 }
49 int main(){
50     int i, j, k, a, b, c;
51     while(scanf("%d %d", &m, &n)==2){
52         CLR(g, 0); CLR(flow, 0);
53         for(i = 0; i < m; ++i){
54             scanf("%d %d %d", &a, &b, &c);
55             g[a][b] += c;//可能有重边
56         }
57         int max_flow = Edmonds_Karp(1, n);
58         printf("%d\n", max_flow);
59     }
60     return 0;
61 }

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