Marriage Match IV(最短路+网络流)

Marriage Match IV

http://acm.hdu.edu.cn/showproblem.php?pid=3416

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6081    Accepted Submission(s): 1766

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

 

 

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define maxn 100005
#define MAXN 10005
#define mem(a,b) memset(a,b,sizeof(a))
const int N=200005;
const int M=200005;
const int INF=0x3f3f3f3f;
using namespace std;
inline int read()
{
    char ch=' ';
    int ans=0;
    while(ch<'0' || ch>'9')
        ch=getchar();
    while(ch<='9' && ch>='0')
    {
        ans=ans*10+ch-'0';
        ch=getchar();
    }
    return ans;
}
int n;
vector<pair<int,int> >ve[maxn];
int dis[maxn],dis1[maxn],dis2[maxn],vis[maxn];
int a[maxn],b[maxn],c[maxn];
struct sair{
    int pos,len;
    friend bool operator<(sair a,sair b){
        return a.len>b.len;
    }
};


void Dijstra(int s){
    for(int i=0;i<=n;i++){
        dis[i]=INF;
        vis[i]=0;
    }
    priority_queue<sair>Q;
    sair tmp;
    tmp.pos=s,tmp.len=0;
    Q.push(tmp);
    dis[s]=0;
    int now,Ne,len;
    while(!Q.empty()){
        tmp=Q.top();
        Q.pop();
        now=tmp.pos;
        if(!vis[now]){
            vis[now]=1;
            for(int i=0;i<ve[now].size();i++){
                Ne=ve[now][i].first;
                len=ve[now][i].second;
                if(dis[Ne]>dis[now]+len){
                    dis[Ne]=dis[now]+len;
                    tmp.pos=Ne;
                    tmp.len=dis[Ne];
                    Q.push(tmp);
                }
            }
        }
    }
}


struct Edge{
    int v,next;
    int cap,flow;
}edge[MAXN*20];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
int cnt=0;//实际存储总边数
void isap_init()
{
    cnt=0;
    memset(pre,-1,sizeof(pre));
}
void isap_add(int u,int v,int w)//加边
{
    edge[cnt].v=v;
    edge[cnt].cap=w;
    edge[cnt].flow=0;
    edge[cnt].next=pre[u];
    pre[u]=cnt++;
}
void add(int u,int v,int w){
    isap_add(u,v,1);
    isap_add(v,u,0);
}
bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    dep[t]=0;
    queue<int>q;
    while(!q.empty())
    q.pop();
    q.push(t);//从汇点开始反向建层次图
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是从汇点反向bfs，但应该判断正向弧的余量
            {
                dep[v]=dep[u]+1;
                gap[dep[v]]++;
                q.push(v);
                //if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
                //break;
            }
        }
    }
    return dep[s]!=-1;
}
int isap(int s,int t)
{
    if(!bfs(s,t))
    return 0;
    memcpy(cur,pre,sizeof(pre));
    //for(int i=1;i<=n;i++)
    //cout<<"cur "<<cur[i]<<endl;
    int u=s;
    path[u]=-1;
    int ans=0;
    while(dep[s]<n)//迭代寻找增广路
    {
        if(u==t)
        {
            int f=INF;
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路
                f=min(f,edge[i].cap-edge[i].flow);
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])
            {
                edge[i].flow+=f;
                edge[i^1].flow-=f;
            }
            ans+=f;
            u=s;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
            {
                cur[u]=path[v]=i;//当前弧优化
                flag=true;
                break;
            }
        }
        if(flag)
        {
            u=v;
            continue;
        }
        int x=n;
        if(!(--gap[dep[u]]))return ans;//gap优化
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
            {
                x=dep[edge[i].v];
                cur[u]=i;//常数优化
            }
        }
        dep[u]=x+1;
        gap[dep[u]]++;
        if(u!=s)//当前点没有增广路则后退一个点
        u=edge[path[u]^1].v;
     }
     return ans;
}

int main(){
    int T;
    T=read();
    while(T--){
        int m;
        n=read();
        m=read();
        for(int i=1;i<=n;i++){
            ve[i].clear();
        }
        for(int i=1;i<=m;i++){
            a[i]=read();
            b[i]=read();
            c[i]=read();
        }
        int s,t;
        s=read();
        t=read();
        for(int i=1;i<=m;i++){
             if(a[i]!=b[i])
            ve[a[i]].push_back(make_pair(b[i],c[i]));
        }
        Dijstra(s);
        memcpy(dis1,dis,sizeof(dis));
        for(int i=1;i<=n;i++){
            ve[i].clear();
        }
        for(int i=1;i<=m;i++){
            if(a[i]!=b[i])
            ve[b[i]].push_back(make_pair(a[i],c[i]));
        }
        Dijstra(t);
        memcpy(dis2,dis,sizeof(dis));
        isap_init();
        for(int i=1;i<=m;i++){
            if(a[i]!=b[i]&&dis1[a[i]]+dis2[b[i]]+c[i]==dis1[t]){
                add(a[i],b[i],1);
            }
        }
        int ans=0;
        ans=isap(s,t);
        printf("%d\n",ans);
    }
}