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(中等) HDU 3416 Marriage Match IV,SPFA+SAP。

时间:2015-03-14 23:08:52      阅读:188      评论:0      收藏:0      [点我收藏+]

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  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?
    
  题意就是求在最短路的基础上有几条路可以到达,但是每条路之间边彼此不能重合。。。
  然后就是先求出最短路来,然后把所以 lowcost[v]==lowcost[u]+cost[u][v] 的边留下,然后再求最大流就好了。。。。。。
 
代码如下:
技术分享
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
    
using namespace std;

const int MaxN=5010;
const int MaxM=500005;
const int INF=10e8;

namespace first
{

struct Edge
{
    int to,next,cost;
};

Edge E[MaxM];
int head[MaxN],Ecou;
int vis[MaxN];

void init(int N)
{
    Ecou=0;

    for(int i=1;i<=N;++i)
    {
        head[i]=-1;
        vis[i]=0;
    }
}

void addEdge(int u,int v,int c)
{
    E[Ecou].to=v;
    E[Ecou].cost=c;
    E[Ecou].next=head[u];
    head[u]=Ecou++;
}

void SPFA(int lowcost[],int N,int start)
{
    queue <int> que;
    int u,v,c;

    for(int i=1;i<=N;++i)
        lowcost[i]=INF;
    lowcost[start]=0;

    que.push(start);
    vis[start]=1;

    while(!que.empty())
    {
        u=que.front();
        que.pop();

        vis[u]=0;

        for(int i=head[u];i!=-1;i=E[i].next)
        {
            v=E[i].to;
            c=E[i].cost;

            if(lowcost[v]>lowcost[u]+c)
            {
                lowcost[v]=lowcost[u]+c;

                if(!vis[v])
                {
                    que.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}

}

namespace second
{

struct Edge
{
    int to,next,cap,flow;
};

Edge E[MaxM];
int Ecou,head[MaxN];
int gap[MaxN],dis[MaxN],pre[MaxN],cur[MaxN];
int S,T;

void init(int N,int _S,int _T)
{
    S=_S;
    T=_T;
    Ecou=0;

    for(int i=1;i<=N;++i)
    {
        head[i]=-1;
        gap[i]=dis[i]=0;
    }
}

void addEdge(int u,int v,int c,int rc=0)
{
    E[Ecou].to=v;
    E[Ecou].cap=c;
    E[Ecou].flow=0;
    E[Ecou].next=head[u];
    head[u]=Ecou++;

    E[Ecou].to=u;
    E[Ecou].cap=rc;
    E[Ecou].flow=0;
    E[Ecou].next=head[v];
    head[v]=Ecou++;
}

void update(int remm)
{
    int u=T;

    while(u!=S)
    {
        E[pre[u]].flow+=remm;
        E[pre[u]^1].flow-=remm;
        u=E[pre[u]^1].to;
    }
}

int SAP(int N)
{
    for(int i=1;i<=N;++i)
        cur[i]=head[i];

    int u,v,ret=0,remm=INF,mindis;

    u=S;
    pre[S]=-1;
    gap[0]=N;

    while(dis[S]<N)
    {
        loop:
        for(int i=cur[u];i!=-1;i=E[i].next)
        {
            v=E[i].to;
            
            if(E[i].cap-E[i].flow && dis[u]==dis[v]+1)
            {
                pre[v]=i;
                cur[u]=i;
                u=v;

                if(u==T)
                {
                    for(int i=pre[u];i!=-1;i=pre[E[i^1].to])
                        remm=min(remm,E[i].cap-E[i].flow);

                    ret+=remm;
                    update(remm);
                    u=S;
                    remm=INF;
                }

                goto loop;
            }
        }

        mindis=N-1;
        for(int i=head[u];i!=-1;i=E[i].next)
            if(E[i].cap-E[i].flow && mindis>dis[E[i].to])
            {
                cur[u]=i;
                mindis=dis[E[i].to];
            }

        if(--gap[dis[u]]==0)
            break;

        dis[u]=mindis+1;

        ++gap[dis[u]];

        if(u!=S)
            u=E[pre[u]^1].to;
    }

    return ret;
}

}

int lowcost[MaxN];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    
    int T;
    int N,M;
    int A,B;
    int a,b,c;

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d %d",&N,&M);

        first::init(N);

        {
            using namespace first;

            while(M--)
            {
                scanf("%d %d %d",&a,&b,&c);

                addEdge(a,b,c);
            }

            scanf("%d %d",&A,&B);

            SPFA(lowcost,N,A);

            second::init(N,A,B);

            for(int u=1;u<=N;++u)
                for(int i=head[u];i!=-1;i=E[i].next)
                    if(lowcost[E[i].to]==lowcost[u]+E[i].cost)
                        second::addEdge(u,E[i].to,1);
        }

        {
            using namespace second;

            printf("%d\n",SAP(N));
        }
    }

    return 0;
}
View Code

 

(中等) HDU 3416 Marriage Match IV,SPFA+SAP。

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原文地址:http://www.cnblogs.com/whywhy/p/4338403.html

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