Flow Problem(最大流模板)

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 17963    Accepted Submission(s): 8464

 

Problem Description

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.

Input

The 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)

Output

For 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

Author

HyperHexagon

Source

HyperHexagon's Summer Gift (Original tasks)

题意：n 个点 m 条有向边，给出 m 条边的 u，v ，cap 问求最大流，算是模板题

练习了Dinic算法

  1 # include 
  2 using namespace std;
  3 # define eps 1e-8
  4 # define INF 0x3f3f3f3f
  5 # define pi  acos(-1.0)
  6 # define MXN  105
  7 # define MXM  1005
  8 
  9 struct Edge{
 10     int from, to, flow, cap;
 11 }edges[MXM*2];              //有向边数
 12 
 13 struct Dinic{
 14     int n, m, s ,t ,idx;    //点，边，源点，汇点，边数
 15     vector G[MXN];     //记录边
 16     int vis[MXN];           //BFS用
 17     int dis[MXN];           //层次
 18     int cur[MXN];           //考虑到哪条弧
 19 
 20     void Init(){
 21         idx=0;
 22         for (int i=1;i<=n;i++) G[i].clear();    //有附加点时要注意
 23     }
 24     
 25     void Addedge(int u,int v,int c){
 26         edges[idx++] = (Edge){u,v,0,c};
 27         edges[idx++] = (Edge){v,u,0,0};
 28         G[u].push_back(idx-2);
 29         G[v].push_back(idx-1);
 30     }
 31 
 32     int BFS()
 33     {
 34         memset(vis,0,sizeof(vis));
 35         queue Q;
 36         Q.push(s);
 37         dis[s] = 0, vis[s] = 1;
 38         while(!Q.empty())
 39         {
 40             int u = Q.front(); Q.pop();
 41             for (int i=0;i<(int)G[u].size();i++)
 42             {
 43                 Edge &e = edges[G[u][i]];
 44                 if (!vis[e.to]&&e.cap>e.flow)
 45                 {
 46                     dis[e.to]=dis[u]+1;
 47                     vis[e.to]=1;
 48                     Q.push(e.to);
 49                 }
 50             }
 51         }
 52         return vis[t];
 53     }
 54 
 55     int DFS(int x,int a)
 56     {
 57         if (x==t||a==0) return a;
 58         int flow = 0, temp;
 59         for (int &i=cur[x];i<(int)G[x].size();i++)
 60         {
 61             Edge &e = edges[G[x][i]];
 62             if (dis[e.to] == dis[x]+1 && (temp=DFS(e.to, min(a, e.cap-e.flow)))>0)
 63             {
 64                 e.flow+=temp;
 65                 edges[G[x][i]^1].flow-=temp;
 66                 flow += temp;
 67                 a-=temp;
 68                 if (a==0) break;
 69             }
 70         }
 71         return flow;
 72     }
 73 
 74     int MaxFlow(int s,int t)
 75     {
 76         this->s = s, this->t = t;
 77         int flow=0;
 78         while(BFS()){
 79             memset(cur,0,sizeof(cur));
 80             flow+=DFS(s,INF);
 81         }
 82         return flow;
 83     }
 84 }F;
 85 
 86 int main()
 87 {
 88     int T;
 89     scanf("%d",&T);
 90     for (int cas=1;cas<=T;cas++)
 91     {
 92         scanf("%d%d",&F.n,&F.m);
 93         F.Init();
 94         for (int i=1;i<=F.m;i++)
 95         {
 96             int u,v,c;
 97             scanf("%d%d%d",&u,&v,&c);
 98             F.Addedge(u,v,c);
 99         }
100         printf("Case %d: %d\n",cas,F.MaxFlow(1,F.n));
101     }
102     return 0;
103 }