HDU 3416 Marriage Match IV

Marriage Match IV

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3416
64-bit integer IO format: %I64d Java class name: Main

Do not sincere non-interference。
Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it's said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.

So, under a good RP, starvae may have many chances to get to city B. But he don't know how many chances at most he can make a data with the girl he likes . Could you help starvae?

Input

The first line is an integer T indicating the case number.(1<=T<=65)
For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it's distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.

Output

Output a line with a integer, means the chances starvae can get at most.

Sample Input

Sample Output

2
1
1

Source

HDOJ Monthly Contest – 2010.06.05

解题：直接把所有的最短路拿出来跑最小割

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 const int INF = 0x3f3f3f3f;
  4 const int maxn = 1010;
  5 struct arc{
  6     int to,w,next;
  7     arc(int x = 0,int y = 0,int z = -1){
  8         to = x;
  9         w = y;
 10         next = z;
 11     }
 12 }e[1000005];
 13 int head[maxn],hd[maxn],d[maxn],gap[maxn],tot,n,m,S,T;
 14 bool in[maxn] = {};
 15 void add(int head[maxn],int u,int v,int wa,int wb){
 16     e[tot] = arc(v,wa,head[u]);
 17     head[u] = tot++;
 18     e[tot] = arc(u,wb,head[v]);
 19     head[v] = tot++;
 20 }
 21 void spfa(){
 22     queue<int>q;
 23     memset(d,0x3f,sizeof d);
 24     d[S] = 0;
 25     q.push(S);
 26     while(!q.empty()){
 27         int u = q.front();
 28         q.pop();
 29         in[u] = false;
 30         for(int i = hd[u]; ~i; i = e[i].next){
 31             if(d[e[i].to] > d[u] + e[i].w){
 32                 d[e[i].to] = d[u] + e[i].w;
 33                 if(!in[e[i].to]){
 34                     in[e[i].to] = true;
 35                     q.push(e[i].to);
 36                 }
 37             }
 38         }
 39     }
 40     for(int i = 1; i <= n; ++i){
 41         for(int j = hd[i]; ~j; j = e[j].next){
 42             if(d[e[j].to] == d[i] + e[j].w)
 43                 add(head,i,e[j].to,1,0);
 44         }
 45     }
 46 }
 47 int dfs(int u,int low){
 48     if(u == T) return low;
 49     int tmp = 0,minH = n - 1;
 50     for(int i = head[u]; ~i; i = e[i].next){
 51         if(e[i].w){
 52             if(d[e[i].to] + 1 == d[u]){
 53                 int a = dfs(e[i].to,min(low,e[i].w));
 54                 e[i].w -= a;
 55                 e[i^1].w += a;
 56                 tmp += a;
 57                 low -= a;
 58                 if(!low) break;
 59                 if(d[S] >= n) return tmp;
 60             }
 61             if(e[i].w) minH = min(minH,d[e[i].to]);
 62         }
 63     }
 64     if(!tmp){
 65         if(--gap[d[u]] == 0) d[S] = n;
 66         ++gap[d[u] = minH + 1];
 67     }
 68     return tmp;
 69 }
 70 void bfs(){
 71     queue<int>q;
 72     memset(d,-1,sizeof d);
 73     memset(gap,0,sizeof gap);
 74     q.push(T);
 75     d[T] = 0;
 76     while(!q.empty()){
 77         int u = q.front();
 78         q.pop();
 79         ++gap[d[u]];
 80         for(int i = head[u]; ~i; i = e[i].next){
 81             if(d[e[i].to] == -1){
 82                 d[e[i].to] = d[u] + 1;
 83                 q.push(e[i].to);
 84             }
 85         }
 86     }
 87 }
 88 int sap(int ret = 0){
 89     bfs();
 90     while(d[S] < n) ret += dfs(S,INF);
 91     return ret;
 92 }
 93 int main(){
 94     int kase,u,v,w;
 95     scanf("%d",&kase);
 96     while(kase--){
 97         scanf("%d%d",&n,&m);
 98         memset(head,-1,sizeof head);
 99         memset(hd,-1,sizeof hd);
100         for(int i = tot = 0; i < m; ++i){
101             scanf("%d%d%d",&u,&v,&w);
102             if(u == v) continue;
103             add(hd,u,v,w,INF);
104         }
105         scanf("%d%d",&S,&T);
106         spfa();
107         printf("%d
",sap());
108     }
109     return 0;
110 }

View Code