Flow Problem(最大流)

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.

InputThe 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)OutputFor 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

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <queue>
 4 #include <algorithm>
 5 using namespace std;
 6 #define inf 0x3f3f3f3f
 7 const int maxn=1005;
 8 struct node{
 9     int to,nex,cap;
10 }e[maxn<<1];
11 int deep[maxn],head[maxn],vis[maxn];
12 int t,n,m,len;
13 void init()
14 {
15     len=0;
16     memset(deep,0,sizeof(deep));
17     memset(e,0,sizeof(e));
18     memset(head,-1,sizeof(head));
19 }
20 void add(int u,int v,int w)
21 {
22     e[len].to=v;
23     e[len].cap=w;
24     e[len].nex=head[u];
25     head[u]=len++;
26 }
27 bool bfs(int s,int t)
28 {
29     memset(vis,0,sizeof(vis));
30     queue<int> q;
31     while(!q.empty()) q.pop();
32     deep[s]=0;
33     vis[s]=1;
34     deep[t]=-1;
35     q.push(s);
36     while(!q.empty())
37     {
38         int u=q.front(); q.pop();
39         for(int i=head[u];i!=-1;i=e[i].nex)
40         {
41             int to=e[i].to;
42             if(e[i].cap<=0||vis[to]) continue;
43             vis[to]=1;
44             deep[to]=deep[u]+1;
45             q.push(to);
46         }
47     }
48     if(deep[t]==-1) return false;
49     else return true;
50 }
51 int dfs(int u,int t,int flow)
52 {
53     int res=0;
54     if(u==t||flow==0) return flow;
55     for(int i=head[u];i!=-1;i=e[i].nex)
56     {
57         int to=e[i].to;
58         if(deep[to]==deep[u]+1)
59         {
60             int f=dfs(to,t,min(flow,e[i].cap)); 
61             if(f<=0) continue;
62             e[i].cap-=f;
63             e[i^1].cap+=f;
64             flow-=f;
65             res+=f;
66             if(flow!=0) break; 
67         }
68     }
69     return res;
70 }
71 int Dinic(int s,int t)
72 {
73     int ans=0;
74     while(bfs(s,t))
75     {
76         ans+=dfs(s,t,inf);
77     }
78     return ans;
79 }
80 int main()
81 {
82     scanf("%d",&t);
83     for(int j=1;j<=t;j++)
84     {
85         init();
86         scanf("%d%d",&n,&m);
87         for(int i=1;i<=m;i++)
88         {
89             int u,v,w;
90             scanf("%d%d%d",&u,&v,&w);
91             add(u,v,w);
92             add(v,u,0);
93         }
94         printf("Case %d: %d\n",j,Dinic(1,n));
95     }
96 }