POJ 1273 Drainage Ditches (网络流Dinic模板)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <string>
 7 #include <cstring>
 8 using namespace std;
 9 const int inf = 0x3f3f3f3f;
10 const int maxn = 220;
11 int c[maxn][maxn];
12 int dep[maxn];
13 int n,m;
14 void pt()
15 {
16     for (int i=1;i<=n;++i){
17             for (int j=1;j<=n;++j)
18                 printf("%d ",c[i][j]);
19             printf("\n");
20         }
21     printf("==================\n");
22 }
23 int bfs (int s,int t)
24 {
25     memset(dep,-1,sizeof dep);
26     queue<int> q;
27     while (!q.empty()) q.pop();
28     dep[s] = 0;
29     q.push(s);
30     while (!q.empty()){
31         int u = q.front();
32         q.pop();
33         for (int v=1;v<=n;++v){
34             if (c[u][v]>0&&dep[v]==-1){//能到达该节点的条件是这条边有流量,而且这个点没有被访问
35                 dep[v] = dep[u]+1;
36                 q.push(v);
37             }
38         }
39     }
40     return dep[t]!=-1;
41 }
42 int dfs (int u,int mi,int t)
43 {
44     if (u==t)
45         return mi;
46     int tmp;
47     for (int v=1;v<=n;++v){
48         if (c[u][v]>0&&dep[v]==dep[u]+1&&(tmp=dfs(v,min(mi,c[u][v]),t))){//下一节点的深度是当前节点+1
49             c[u][v]-=tmp;
50             c[v][u]+=tmp;
51             return tmp;
52         }
53     }
54     return 0;//别忘写返回0!!!
55 }
56 int dinic ()
57 {
58     int ans = 0;
59     int tmp;
60     while (bfs(1,n)){//每次按照深度建立分层图,这样每次dfs的时候下一节点的深度是当前节点+1
61         while (1){
62             tmp = dfs(1,inf,n);
63             //printf("%d\n",tmp);
64             if (tmp==0)
65                 break;
66             //pt();
67             ans+=tmp;
68         }
69     }
70     return ans;
71 }
72 int main()
73 {
74     //freopen("de.txt","r",stdin);
75     while (~scanf("%d%d",&m,&n)){
76         memset(c,0,sizeof c);
77         for (int i=0;i<m;++i){
78             int u,v,cap;
79             scanf("%d%d%d",&u,&v,&cap);
80             c[u][v]+=cap;
81         }
82         printf("%d\n",dinic());
83     }
84     return 0;
85 }