HDU3416(KB11-O spfa+最大流)

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4381    Accepted Submission(s): 1310

Problem Description

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

 

 

Author

starvae@HDU

 

 

Source

HDOJ Monthly Contest – 2010.06.05

 

只有最短路上的边才加入网络，容量都为1，跑最大流

//2017-08-25
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 5010;
const int M = 500000;
const int INF = 0x3f3f3f3f;
int head[N], tot;
struct Edge{
    int next, to, w;
}edge[M];

void add_edge(int u, int v, int w){
    edge[tot].w = w;
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

struct Dinic{
    int level[N], S, T;
    void init(){
        tot = 0;
        memset(head, -1, sizeof(head));
    }
    void set_s_t(int _S, int _T){
        S = _S;
        T = _T;
    }
    bool bfs(){
        queue<int> que;
        memset(level, -1, sizeof(level));
        level[S] = 0;
        que.push(S);
        while(!que.empty()){
            int u = que.front();
            que.pop();
            for(int i = head[u]; i != -1; i = edge[i].next){
                int v = edge[i].to;
                int w = edge[i].w;
                if(level[v] == -1 && w > 0){
                    level[v] = level[u]+1;
                    que.push(v);
                }
            }
        }
        return level[T] != -1;
    }
    int dfs(int u, int flow){
        if(u == T)return flow;
        int ans = 0, fw;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            if(!w || level[v] != level[u]+1)
              continue;
            fw = dfs(v, min(flow-ans, w));
            ans += fw;
            edge[i].w -= fw;
            edge[i^1].w += fw;
            if(ans == flow)return ans;
        }
        if(ans == 0)level[u] = 0;
        return ans;
    }
    int maxflow(){
        int flow = 0;
        while(bfs())
          flow += dfs(S, INF);
        return flow;    
    }
}dinic;

//dis[0][u]表示从起点到u的最短距离，dis[1][u]表示从终点到u的最短距离，cnt[u]记录u入队次数，判负环用
int dis[2][N], cnt[N];
bool vis[N];
bool spfa(int s, int n, int op){
    memset(vis, false, sizeof(vis));
    memset(dis[op], INF, sizeof(dis));
    memset(cnt, 0, sizeof(cnt));
    vis[s] = true;
    dis[op][s] = 0;
    cnt[s] = 1;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].to;
            int w = edge[i].w;
            if(dis[op][v] > dis[op][u] + w){
                dis[op][v] = dis[op][u] + w;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                    if(++cnt[v] > n)return false;
                }
            }
        }
    }
    return true;
}

struct Line{
    int u, v, w;
}line[M];

int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("inputO.txt", "r", stdin);
    int T, n, m, A, B;
    cin>>T;
    while(T--){
        cin>>n>>m;
        int u, v, w;
        for(int i = 0; i < m; i++)
          cin>>line[i].u>>line[i].v>>line[i].w;
        cin>>A>>B;
        dinic.init();
        for(int i = 0; i < m; i++)
          add_edge(line[i].u, line[i].v, line[i].w);
        spfa(A, n, 0);//求从起点开始到各个点的最短路
        dinic.init();
        for(int i = 0; i < m; i++)
          add_edge(line[i].v, line[i].u, line[i].w);
        spfa(B, n, 1);//求从终点开始到各个点的最短路
        dinic.init();
        dinic.set_s_t(A, B);
        for(int i = 0; i < m; i++){
            u = line[i].u;
            v = line[i].v;
            w = line[i].w;
            //只有最短路上的边才加入网络
            if(dis[0][u]+dis[1][v]+w == dis[0][B]){
                add_edge(u, v, 1);
                add_edge(v, u, 0);
            }
        }
        cout<<dinic.maxflow()<<endl;
    }
    return 0;
}