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

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <climits>
  7 #include <vector>
  8 #include <queue>
  9 #include <cstdlib>
 10 #include <string>
 11 #include <set>
 12 #include <stack>
 13 #define LL long long
 14 #define INF 0x3f3f3f3f
 15 using namespace std;
 16 const int maxn = 10010;
 17 struct arc{
 18     int to,flow,next;
 19     arc(int x = 0,int y = 0,int z = -1){
 20         to = x;
 21         flow = y;
 22         next = z;
 23     }
 24 };
 25 arc e[maxn*10];
 26 int head[maxn],d[maxn],cur[maxn],tot,s,t,n;
 27 void add(int u,int v,int flow){
 28     e[tot] = arc(v,flow,head[u]);
 29     head[u] = tot++;
 30     e[tot] = arc(u,0,head[v]);
 31     head[v] = tot++;
 32 }
 33 bool bfs(){
 34     queue<int>q;
 35     for(int i = 0; i <= n; ++i) d[i] = -1;
 36     q.push(s);
 37     d[s] = 1;
 38     while(!q.empty()){
 39         int u = q.front();
 40         q.pop();
 41         for(int i = head[u]; ~i; i = e[i].next){
 42             if(e[i].flow && d[e[i].to] == -1){
 43                 d[e[i].to] = d[u] + 1;
 44                 q.push(e[i].to);
 45             }
 46         }
 47     }
 48     return d[t] > -1;
 49 }
 50 int dfs(int u,int low){
 51     if(u == t) return low;
 52     int tmp = 0,a;
 53     for(int &i = cur[u]; ~i; i = e[i].next){
 54         if(e[i].flow && d[e[i].to] == d[u] + 1 && (a = dfs(e[i].to,min(e[i].flow,low)))){
 55             e[i].flow -= a;
 56             e[i^1].flow += a;
 57             tmp += a;
 58             low -= a;
 59             break;
 60         }
 61     }
 62     if(!tmp) d[u] = -1;
 63     return tmp;
 64 }
 65 bool vis[maxn];
 66 int maxcap;
 67 void dfs2(int u){
 68     vis[u] = true;
 69     for(int i = head[u]; ~i; i = e[i].next){
 70         if(!vis[e[i].to]){
 71             maxcap = max(maxcap,e[i^1].flow);
 72             if(e[i^1].flow == 5){
 73                 cout<<u<<" "<<e[i].to<<endl;
 74             }
 75             dfs2(e[i].to);
 76         }
 77     }
 78 }
 79 int main(){
 80     int p,cs,m,u,v,cap;
 81     scanf("%d",&p);
 82     while(p--){
 83         scanf("%d %d %d %d %d",&cs,&n,&m,&s,&t);
 84         memset(head,-1,sizeof(head));
 85         for(int i = tot = 0; i < m; ++i){
 86             scanf("%d %d %d",&u,&v,&cap);
 87             add(u,v,cap);
 88         }
 89         int ans = 0,o;
 90         maxcap = 0;
 91         while(bfs()){
 92             memcpy(cur,head,sizeof(head));
 93             o = dfs(s,INF);
 94             ans += o;
 95             maxcap = max(maxcap,o);
 96         }
 97         memset(vis,false,sizeof(vis));
 98         //maxcap = 0;
 99         //dfs2(s);
100         printf("%d %.3f
",cs,ans*1.0/maxcap);
101         //cout<<ans<<" "<<maxcap<<endl;
102     }
103     return 0;
104 }

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