HDU 4240 Route Redundancy

Route Redundancy

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4240
64-bit integer IO format: %I64d Java class name: Main

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

Sample Output

1 1.667

Source

2011 Greater New York Regional

解题：最大流

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <climits>

 #include <vector>

 #include <queue>

 #include <cstdlib>

 #include <string>

 #include <set>

 #include <stack>

 #define LL long long

 #define INF 0x3f3f3f3f

 using namespace std;

 const int maxn = ;

 struct arc{

     int to,flow,next;

     arc(int x = ,int y = ,int z = -){

         to = x;

         flow = y;

         next = z;

     }

 };

 arc e[maxn*];

 int head[maxn],d[maxn],cur[maxn],tot,s,t,n;

 void add(int u,int v,int flow){

     e[tot] = arc(v,flow,head[u]);

     head[u] = tot++;

     e[tot] = arc(u,,head[v]);

     head[v] = tot++;

 }

 bool bfs(){

     queue<int>q;

     for(int i = ; i <= n; ++i) d[i] = -;

     q.push(s);

     d[s] = ;

     while(!q.empty()){

         int u = q.front();

         q.pop();

         for(int i = head[u]; ~i; i = e[i].next){

             if(e[i].flow && d[e[i].to] == -){

                 d[e[i].to] = d[u] + ;

                 q.push(e[i].to);

             }

         }

     }

     return d[t] > -;

 }

 int dfs(int u,int low){

     if(u == t) return low;

     int tmp = ,a;

     for(int &i = cur[u]; ~i; i = e[i].next){

         if(e[i].flow && d[e[i].to] == d[u] +  && (a = dfs(e[i].to,min(e[i].flow,low)))){

             e[i].flow -= a;

             e[i^].flow += a;

             tmp += a;

             low -= a;

             break;

         }

     }

     if(!tmp) d[u] = -;

     return tmp;

 }

 bool vis[maxn];

 int maxcap;

 void dfs2(int u){

     vis[u] = true;

     for(int i = head[u]; ~i; i = e[i].next){

         if(!vis[e[i].to]){

             maxcap = max(maxcap,e[i^].flow);

             if(e[i^].flow == ){

                 cout<<u<<" "<<e[i].to<<endl;

             }

             dfs2(e[i].to);

         }

     }

 }

 int main(){

     int p,cs,m,u,v,cap;

     scanf("%d",&p);

     while(p--){

         scanf("%d %d %d %d %d",&cs,&n,&m,&s,&t);

         memset(head,-,sizeof(head));

         for(int i = tot = ; i < m; ++i){

             scanf("%d %d %d",&u,&v,&cap);

             add(u,v,cap);

         }

         int ans = ,o;

         maxcap = ;

         while(bfs()){

             memcpy(cur,head,sizeof(head));

             o = dfs(s,INF);

             ans += o;

             maxcap = max(maxcap,o);

         }

         memset(vis,false,sizeof(vis));

         //maxcap = 0;

         //dfs2(s);

         printf("%d %.3f\n",cs,ans*1.0/maxcap);

         //cout<<ans<<" "<<maxcap<<endl;

     }

     return ;

 }

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