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

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

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 ;
 }