hdu 4240(最大流+最大流量的路)

Route Redundancy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 625    Accepted Submission(s): 367

Problem Description

A city is made up exclusively of one-way steets.each street in the city has a capacity,which is the minimum of the capcities of the streets along that route.

The redundancy ratio from point A to point B is the ratio of the maximum number of cars that can get from point A to point B in an hour using all routes simultaneously,to the maximum number of cars thar can get from point A to point B in an hour using one route.The minimum redundancy ratio is the number of capacity of the single route with the laegest capacity.

 

Input

The first line of input contains asingle integer P,(1<=P<=1000),which is the number of data sets that follow.Each data set consists of several lines and represents a directed graph with positive integer weights.

The first line of each data set contains five apace separatde integers.The first integer,D is the data set number. The second integer,N(2<=N<=1000),is the number of nodes inthe graph. The thied integer,E,(E>=1),is the number of edges in the graph. The fourth integer,A,(0<=A<N),is the index of point A.The fifth integer,B,(o<=B<N,A!=B),is the index of point B.

The remaining E lines desceibe each edge. Each line contains three space separated in tegers.The First integer,U(0<=U<N),is the index of node U. The second integer,V(0<=v<N,V!=U),is the node V.The third integer,W (1<=W<=1000),is th capacity (weight) of path from U to V.

 

Output

For each data set there is one line of output.It contains the date set number(N) follow by a single space, followed by a floating-point value which is the minimum redundancy ratio to 3 digits after the decimal point.

 

Sample Input

1 1 7 11 0 6 0 1 3 0 3 3 1 2 4 2 0 3 2 3 1 2 4 2 3 4 2 3 5 6 4 1 1 4 6 1 5 6 9

 

Sample Output

1 1.667

题意:求解 最大流 / 图里面最大流量的那一条路 是多少??

http://www.cnblogs.com/yezekun/p/3925768.html正确题解提供者。

题解:此题正确解法不是在dfs时找增广路时更新,那样的话会出两个问题.所以需要先要预处理出最大流量的那条路.然后再求最大流。这样才是正确的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
const int N = 1005;
const int INF = 999999999;
struct Edge{
    int v,w,next;
}edge[N*N];
int head[N];
int level[N];
int tot,max_increase;
void init()
{
    memset(head,-1,sizeof(head));
    tot=0;
}
void addEdge(int u,int v,int w,int &k)
{
    edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++;
    edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++;
}
int BFS(int src,int des)
{
    queue<int>q;
    memset(level,0,sizeof(level));
    level[src]=1;
    q.push(src);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(u==des) return 1;
        for(int k = head[u]; k!=-1; k=edge[k].next)
        {
            int v = edge[k].v;
            int w = edge[k].w;
            if(level[v]==0&&w!=0)
            {
                level[v]=level[u]+1;
                q.push(v);
            }
        }
    }
    return -1;
}
int dfs(int u,int des,int increaseRoad){
    if(u==des||increaseRoad==0) {
        return increaseRoad;
    }
    int ret=0;
    for(int k=head[u];k!=-1;k=edge[k].next){
        int v = edge[k].v,w=edge[k].w;
        if(level[v]==level[u]+1&&w!=0){
            int MIN = min(increaseRoad-ret,w);
            w = dfs(v,des,MIN);
            if(w > 0)
            {
                edge[k].w -=w;
                edge[k^1].w+=w;
                ret+=w;
                if(ret==increaseRoad){
                    return ret;
                }
            }
            else level[v] = -1;
            if(increaseRoad==0) break;
        }
    }
    if(ret==0) level[u]=-1;
    return ret;
}
int Dinic(int src,int des)
{
    int ans = 0;
    while(BFS(src,des)!=-1) ans+=dfs(src,des,INF);
    return ans;
}
int d,n,m,src,des;
bool vis[N];
void dfs1(int u,int ans){
    if(u==des){
        max_increase = max(max_increase,ans);
        return ;
    }
    for(int k=head[u];k!=-1;k=edge[k].next){
        int v = edge[k].v,w = edge[k].w;
        if(!vis[v]){
            vis[v] = true;
            dfs1(v,min(w,ans)); ///最大流量由最小容量边决定
            vis[v] = false;
        }
    }
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        init();
        max_increase = -1;
        scanf("%d%d%d%d%d",&d,&n,&m,&src,&des);
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addEdge(u,v,w,tot);
        }
        memset(vis,false,sizeof(vis));
        dfs1(src,99999999); ///deal
        int max_flow = Dinic(src,des);
        printf("%d %.3lf
",d,max_flow*1.0/max_increase);
    }
    return 0;
}