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

3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2

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

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1010;
struct arc{
    int to,w,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        w = y;
        next = z;
    }
}e[1000005];
int head[maxn],hd[maxn],d[maxn],gap[maxn],tot,n,m,S,T;
bool in[maxn] = {};
void add(int head[maxn],int u,int v,int wa,int wb){
    e[tot] = arc(v,wa,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,wb,head[v]);
    head[v] = tot++;
}
void spfa(){
    queue<int>q;
    memset(d,0x3f,sizeof d);
    d[S] = 0;
    q.push(S);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        in[u] = false;
        for(int i = hd[u]; ~i; i = e[i].next){
            if(d[e[i].to] > d[u] + e[i].w){
                d[e[i].to] = d[u] + e[i].w;
                if(!in[e[i].to]){
                    in[e[i].to] = true;
                    q.push(e[i].to);
                }
            }
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = hd[i]; ~j; j = e[j].next){
            if(d[e[j].to] == d[i] + e[j].w)
                add(head,i,e[j].to,1,0);
        }
    }
}
int dfs(int u,int low){
    if(u == T) return low;
    int tmp = 0,minH = n - 1;
    for(int i = head[u]; ~i; i = e[i].next){
        if(e[i].w){
            if(d[e[i].to] + 1 == d[u]){
                int a = dfs(e[i].to,min(low,e[i].w));
                e[i].w -= a;
                e[i^1].w += a;
                tmp += a;
                low -= a;
                if(!low) break;
                if(d[S] >= n) return tmp;
            }
            if(e[i].w) minH = min(minH,d[e[i].to]);
        }
    }
    if(!tmp){
        if(--gap[d[u]] == 0) d[S] = n;
        ++gap[d[u] = minH + 1];
    }
    return tmp;
}
void bfs(){
    queue<int>q;
    memset(d,-1,sizeof d);
    memset(gap,0,sizeof gap);
    q.push(T);
    d[T] = 0;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        ++gap[d[u]];
        for(int i = head[u]; ~i; i = e[i].next){
            if(d[e[i].to] == -1){
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
    }
}
int sap(int ret = 0){
    bfs();
    while(d[S] < n) ret += dfs(S,INF);
    return ret;
}
int main(){
    int kase,u,v,w;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d",&n,&m);
        memset(head,-1,sizeof head);
        memset(hd,-1,sizeof hd);
        for(int i = tot = 0; i < m; ++i){
            scanf("%d%d%d",&u,&v,&w);
            if(u == v) continue;
            add(hd,u,v,w,INF);
        }
        scanf("%d%d",&S,&T);
        spfa();
        printf("%d
",sap());
    }
    return 0;
}