Flow Problem

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3549

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 19255    Accepted Submission(s): 9046

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

 

 题解：裸地网络流求1到n

#include 
      
       using namespace std;#define INF 100000008const int maxn = 20,maxm = 4005;struct Netflow{    int S,T,tot,head[maxn],dis[maxn]，hh[maxn];    bool vis[maxn];    struct edge{        int to,f,nxt;    }G[maxm];    void init(int s, int t){        tot = 1;        memset(head,0,sizeof(head));        memset(head,0,sizeof(head));        S = s, T = t;    }    void add(int u,int v,int w){        G[++tot].nxt = head[u];        G[tot].to = v;        G[tot].f = w;        head[u] = tot;                G[++tot].nxt = head[v];        G[tot].to = u;        G[tot].f = 0;        head[v] = tot;    }    bool bfs(){        memset(vis,0,sizeof(vis));        memset(dis,0,sizeof(dis));        vis[S] = 1;        queue 
       
         q;        q.push(S);        while(!q.empty()){            int u = q.front();            q.pop();            for(int i = head[u]; i; i = G[i].nxt){                int v = G[i].to;                if(!vis[v] && G[i].f){                    vis[v] = 1;                    dis[v] = dis[u] + 1;                    q.push(v);                }              }        }         return vis[T];    }    int dfs(int u,int v){        if(u == T || !v )return v;        int ret = 0;        for(int i = hh[u]; i; i = G[i].nxt){            int p = G[i].to;            if(G[i].f && dis[p] == dis[u] + 1){                int dd = dfs(p, min(v, G[i].f));                G[i].f -= dd;                G[i^1].f += dd;                v-=dd;                ret += dd;               hh[u] = i;            }        }        if(!ret)dis[u]=-1;        return ret;    }    int dinic(){        int ans = 0;        while(bfs()){            for(int i = S; i <= T; i++)hh[i] = head[i];            ans += dfs(S,INF);        }        return ans;    }    };Netflow Tr;int main(){    int T;    scanf("%d",&T);    for(int k = 1; k <= T; k++){        int n,m;        scanf("%d%d",&n,&m);        Tr.init(1,n);        for(int i = 1; i <= m; i++){            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            Tr.add(u,v,w);        }        printf("Case %d: %d\n",k,Tr.dinic());            }}