hdu3416 Marriage Match IV(最短路+最大流)

最新推荐文章于 2021-05-12 09:08:10 发布

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最新推荐文章于 2021-05-12 09:08:10 发布

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Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4743 Accepted Submission(s): 1420

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

Sample Output

2
1
1

题意：有n个城市，m条边，每条边表示u到v花费时间w，问从s到t花费时间最小的情况下有多少种方案

问花费时间最小，我们可以先求出以s为到t的最短时间然后把无用边删去，再求一下最大流即可

关于删边:记最小花费为cost

可以先以s为源点，求最短路，此时数组记为dis1

再以t为源点，求最短路，此时数组记为dis2

当dis1[u]+e[u][i]+dis2[i]==cost时，这条边为可用边

最后这题比较卡时间（还是kuangbin的sap板子快啊）

#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
const int N=2000+20;
const int M=2000000+20;
int top;
int h[N],pre[N],g[N],first[N],cur[N];//h[i]记录每个节点的高度，pre[i]记录前驱,g[i]表示距离为i个节点数有多少个
struct node
{
    int v,next,cap;
} E[M];
void init()
{
    mem(first,-1);
    top=0;
}
void add_edge(int u,int v,int c)
{
    E[top].v=v;E[top].cap=c;E[top].next=first[u];first[u]=top++;
    E[top].v=u;E[top].cap=0;E[top].next=first[v];first[v]=top++;
}
int sap(int start,int end,int nodenum)
{
    memset(h,0,sizeof(h));
    memset(g,0,sizeof(g));
    memcpy(cur,first,sizeof(first));
    int u=pre[start]=start,maxflow=0,aug=-1;
    g[0]=nodenum;
    while(h[start]<nodenum)
    {
        loop:
        for(int &i=cur[u];i!=-1;i=E[i].next)
        {
            int v=E[i].v;
            if(E[i].cap&&h[u]==h[v]+1)
            {
                if(aug==-1||aug>E[i].cap)
                    aug=E[i].cap;
                pre[v]=u;
                u=v;
                if(v==end)
                {
                    maxflow+=aug;
                    for(u=pre[u];v!=start;v=u,u=pre[u])
                    {
                        E[cur[u]].cap-=aug;
                        E[cur[u]^1].cap+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        }
        int mindis=nodenum;
        for(int i=first[u];i!=-1;i=E[i].next)
        {
            int v=E[i].v;
            if(E[i].cap&&mindis>h[v])
            {
                cur[u]=i;
                mindis=h[v];
            }
        }
        if((--g[h[u]])==0)break;
        g[h[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}

struct point
{
    int x,w;
    point() {}
    point(int tx,int tw)
    {
        x=tx,w=tw;
    };
    bool friend operator <(point a,point b)
    {
        return  a.w>b.w;
    }

};
struct edge
{
    int v,w,next;
} e[M];
int kcase,m,k,n,x[M],y[M],w[M],z,s,t,head[N],tt,c[N],vis[N];
struct solve
{
    int dis[N];
    void init()
    {
        mem(head,-1);
        tt=0;
    }
    void add_edge(int u,int v,int w)
    {
            e[tt].v=v;
            e[tt].w=w;
            e[tt].next=head[u];
            head[u]=tt++;
    }
    bool spfa(int s)
    {
        priority_queue<point>q;
        for(int i=1; i<=n; i++)
            c[i]=0,vis[i]=0,dis[i]=INF;
        q.push(point(s,0));
        vis[s]=1;
        dis[s]=0;
        c[s]++;
        while(!q.empty())
        {
            point u=q.top();
            q.pop();
            vis[u.x]=false;
            for(int i=head[u.x];i!=-1;i=e[i].next)
            {

                int v=e[i].v;
                if(dis[v]>dis[u.x]+e[i].w)
                {
                    dis[v]=dis[u.x]+e[i].w;
                    if(!vis[v])
                    {
                        c[v]++;
                        if(c[v]>n)return false;
                        vis[v]=1;
                        q.push(point(e[i].v,dis[v]));
                    }
                }
            }
        }
        return true;
    }
};
int main()
{
    scanf("%d",&kcase);
    while(kcase--)
    {

        scanf("%d%d",&n,&m);
        solve a,b;
        a.init();
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&x[i],&y[i],&w[i]);
            a.add_edge(x[i],y[i],w[i]);
        }
        scanf("%d%d",&s,&t);
        a.spfa(s);

        b.init();
        for(int i=1; i<=m; i++)
             b.add_edge(y[i],x[i],w[i]);
        b.spfa(t);
        init();
        for(int i=1; i<=m; i++)
            if(x[i]!=y[i]&&a.dis[x[i]]+b.dis[y[i]]+w[i]==a.dis[t])
                add_edge(x[i],y[i],1);
        printf("%d\n",sap(s,t,n));
    }
    return 0;
}