hdu3599 War(最大流)

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War

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1534    Accepted Submission(s): 334

Problem Description

In the war between Anihc and Bandzi, Anihc has won Bangzi. At that time, Lvbin, the king of Anihc, want to start from Bangzi to beat Nebir across the channel between them. He let his army get to there as soon as possible, and there located many islands which can be used to have a break. In order to save time, the king forbid the army getting through the same pass for more than one time, but they can reach the same island for as many as times they want.
Yunfeng, the General of the army, must tell how many optimal ship routes are there to the king as soon as possible, or he will be killed. Now he asks for your help. You must help Yunfeng to save his life.
He tells you that there are N island. The islands are numbered from 1 to N(1 is Bangzi and N is Nebir, others are many islands). And there are many ways, each way contain the islands number U and V and the length W. Please output your answer.

 

Input

The first line in the input file contains a single integer number T means the case number. 
Each case contains a number N (N<=1500) means the number of the islands. 
And then many lines follow. Each line contains three numbers: U V W (W<10000), means that the distance between island U and V is W. The input of ways are terminated by “ 0 0 0 ”.

 

Output

Print the number of the optimal ship routes are there after each case in a line.

 

Sample Input

1
6
1 2 1
3 2 1
3 4 1
1 3 2
4 2 2
4 5 1
5 6 1
4 6 2
0 0 0

 

Sample Output

2

 

Author

alpc92

 

Source

2010 ACM-ICPC Multi-University Training Contest（16）——Host by NUDT

题意：

问有几种最短路的方案。每条边只能经过一次。

分析：

跑一遍最短路，若某条边是最短路的中的边，则连一条对应的容量为1的边，然后dinic搞一下就行

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>
 using namespace std;
 #define XINF INT_MAX
 #define INF 0x3FFFFFFF
 #define MP(X,Y) make_pair(X,Y)
 #define PB(X) push_back(X)
 #define REP(X,N) for(int X=0;X<N;X++)
 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 #define CLR(A,X) memset(A,X,sizeof(A))
 #define IT iterator
 typedef long long ll;
 typedef pair<int,int> PII;
 typedef vector<PII> VII;
 typedef vector<int> VI;
 #define MAXN 1600
 vector<PII> Map[MAXN];

 void init() { REP(i,MAXN) Map[i].clear(); }

 int dis[MAXN];
 void dijkstra(int s)
 {
     REP(i,MAXN){dis[i]=i==s?:INF;}
     int vis[MAXN] = {};
     priority_queue<PII, vector<PII>, greater<PII> > q;
     q.push(MP(,s));
     while(!q.empty())
     {
         PII p = q.top(); q.pop();
         int x = p.second;
         if(vis[x])continue;
         vis[x] = ;
         for(vector<PII>::iterator it = Map[x].begin(); it != Map[x].end(); it++)
         {
             int y = it->first;
             int d = it->second;
             if(!vis[y] && dis[y] > dis[x] + d)
             {
                 dis[y] = dis[x] + d;
                 q.push(MP(dis[y],y));
             }
         }
     }
 }
 struct edge{
     int to,cap,rev;
     edge(int _to,int _cap,int _rev)
     {
         to=_to;
         cap=_cap;
         rev=_rev;
     }
 };
 const int MAX_V=;
 vector<edge>G[MAX_V];
 int iter[MAX_V];
 int level[MAX_V];
 int tot=;
 void add_edge(int from,int to,int cap)
 {
     G[from].PB(edge(to,cap,G[to].size()));
     G[to].PB(edge(from,,G[from].size()-));
 }
 void bfs(int s,int t)
 {
     CLR(level,-);
     queue<int>q;
     level[s]=;
     q.push(s);
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         for(int i=;i<G[u].size();i++)
         {
             edge &e=G[u][i];
             if(e.cap>&&level[e.to]<)
             {
                 level[e.to]=level[u]+;
                 q.push(e.to);
             }
         }
     }
 }
 int dfs(int v,int t,int f)
 {
     if(v==t)return f;
     for(int &i=iter[v];i<G[v].size();i++)
     {
         edge &e=G[v][i];
         if(e.cap>&&level[v]<level[e.to])
         {
             int d=dfs(e.to,t,min(f,e.cap));
             if(d>)
             {
                 e.cap-=d;;
                 G[e.to][e.rev].cap+=d;
                 return d;
             }
         }
     }
     return ;
 }
 int Dinic(int s,int t)
 {
     int flow=;
     for(;;)
     {
         bfs(s,t);
         if(level[t]<)return flow;
         memset(iter,,sizeof(iter));
         int f;
         while((f=dfs(s,t,INF))>)
         {
             flow+=f;
         }
     }
 }

 int main()
 {
     ios::sync_with_stdio(false);
     int t;
     scanf("%d",&t);
     while(t--){
         int n;
         init();
         scanf("%d",&n);
         int u,v,d;
         while(scanf("%d%d%d",&u,&v,&d)&&(u||v||d)){
             u--;
             v--;
             Map[u].PB(MP(v,d));
             Map[v].PB(MP(u,d));
         }
         if(n==){
             printf("0\n");
             continue;
         }
         for(int i=;i<n;i++)G[i].clear();
         dijkstra();
         for(int i=;i<n;i++){
             for(vector<PII>::iterator it = Map[i].begin(); it != Map[i].end(); it++)
             {
                 int y = it->first;
                 int d = it->second;
                 if(dis[i]+d==dis[y]){
                     add_edge(i,y,);
                 }
             }
         }
         printf("%d\n",Dinic(,n-));
     }

     return ;
 }

代码君