hdu 4240 最大流量路径

题意弄了半天:

给出一个有向图,带边权,src,dst. 求出src到dst的最大流,再求出从src到dst流量最大的路径的流量,求它们的比值。

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #include <vector>
  5 #define oo 0x3f3f3f3f
  6 #define maxn 1010
  7 using namespace std;
  8 
  9 struct Edge {
 10     int u, v, f, cap;
 11     Edge( int u, int v, int f, int cap ):u(u),v(v),f(f),cap(cap){}
 12 };
 13 struct Dinic {
 14     int n, src, dst;
 15     vector<Edge> edge;
 16     vector<int> g[maxn];
 17     int dep[maxn], cur[maxn], sig;
 18 
 19     void init( int n, int src, int dst ) {
 20         this->n = n;
 21         this->src = src;
 22         this->dst = dst;
 23         for( int u=1; u<=n; u++ )
 24             g[u].clear();
 25         edge.clear();
 26     }
 27     void add_edge( int u, int v, int f ) {
 28         g[u].push_back( edge.size() );
 29         edge.push_back( Edge(u,v,f,f) );
 30         g[v].push_back( edge.size() );
 31         edge.push_back( Edge(v,u,0,f) );
 32     }
 33     bool bfs() {
 34         queue<int> qu;
 35         memset( dep, 0, sizeof(dep) );
 36         qu.push( src );
 37         dep[src] = 1;
 38         while( !qu.empty() ) {
 39             int u=qu.front();
 40             qu.pop();
 41             for( int t=0; t<g[u].size(); t++ ) {
 42                 Edge &e = edge[g[u][t]];
 43                 if( e.f && !dep[e.v] ) {
 44                     dep[e.v] = dep[e.u]+1;
 45                     qu.push( e.v );
 46                 }
 47             }
 48         }
 49         return dep[dst];
 50     }
 51     int dfs( int u, int a, int minc ) {
 52         if( u==dst || a==0 ) {
 53             sig = max( minc, sig );
 54             return a;
 55         }
 56         int remain=a, past=0, na;
 57         for( int &t=cur[u]; t<g[u].size(); t++ ) {
 58             Edge &e = edge[g[u][t]];
 59             Edge &ve = edge[g[u][t]^1];
 60             if( dep[e.v]==dep[e.u]+1 && (na=dfs(e.v,min(remain,e.f),min(minc,e.cap))) ) {
 61                 remain -= na;
 62                 past += na;
 63                 e.f -= na;
 64                 ve.f += na;
 65                 if( remain==0 ) break;
 66             }
 67         }
 68         return past;
 69     }
 70     void maxflow( int &tot, int &sig ) {
 71         tot = sig = 0;
 72         this->sig = 0;
 73         while( bfs() ) {
 74             memset( cur, 0, sizeof(cur) );
 75             tot += dfs(src,oo,oo);
 76         }
 77         sig = this->sig;
 78     }
 79 };
 80 
 81 int n, m;
 82 Dinic D;
 83 int main() {
 84     int T;
 85     scanf( "%d", &T );
 86     while( T-- ) {
 87         int cas, src, dst;
 88         scanf( "%d%d%d%d%d", &cas, &n, &m, &src, &dst );
 89         src++, dst++;
 90         D.init( n, src, dst );
 91         for( int i=1,u,v,w; i<=m; i++ ) {
 92             scanf( "%d%d%d", &u, &v, &w );
 93             u++, v++;
 94             D.add_edge( u, v, w );
 95         }
 96         int tot, sig;
 97         D.maxflow( tot, sig );
 98         printf( "%d %.3lf
", cas, double(tot)/double(sig) );
 99     }
100 }
View Code
原文地址:https://www.cnblogs.com/idy002/p/4321754.html