hdu5294||2015多校联合第一场1007 最短路+最大流

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http://acm.hdu.edu.cn/showproblem.php?pid=5294

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/**
hdu5294 最短路+最大流
题目大意：给定一个无向图，从起点到终点，只有走最短路，才能在规定时限内到达，问最少去掉几条边使不能到达，最多去掉几条边仍能到达
解题思路：http://blog.sina.com.cn/s/blog_15139f1a10102vnx5.html 和官方题解想的一样
*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
const int oo=1e9;
const int mm=161111;
const int mn=2330;
int node,src,dest,edge;
int ver[mm],flow[mm],_next[mm];
int head[mn],work[mn],dis[mn],q[mn];

void prepare(int _node,int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<=node; ++i)head[i]=-1;
    edge=0;
}

void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,_next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,_next[edge]=head[v],head[v]=edge++;
}

bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i<node; ++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0; l<r; ++l)
        for(i=head[u=q[l]]; i>=0; i=_next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    for(int &i=work[u],v,tmp; i>=0; i=_next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0; i<node; ++i)work[i]=head[i];
        while(delta=Dinic_dfs(src,oo))ret+=delta;
    }
    return ret;
}

///==================================================
const int INF=0x3f3f3f3f;
const int maxm=511111;
const int maxn=2111;

struct EdgeNode
{
    int to;
    int w;
    int next;
};

EdgeNode edges[maxm];
int N,M;
int head1[maxn],edge1;
bool vis[maxn];
queue <int> que;
int dis1[maxn],dis2[maxn];

void addedge1(int u,int v,int c)
{
    edges[edge1].w=c,edges[edge1].to=v,edges[edge1].next=head1[u],head1[u]=edge1++;
}

void init()
{
    memset(head1,-1,sizeof(head1));
    edge1=0;
}

void spfa(int s,int n)//单源最短路(s为起点，n为节点总数)
{
    int u;
    for (int i=0; i<=n; i++)
        dis1[i]=INF;
    memset(vis,0,sizeof(vis));
    while (!que.empty()) que.pop();
    que.push(s);
    vis[s]=true;
    dis1[s]=0;
    while (!que.empty())
    {
        u=que.front();
        que.pop();
        vis[u]=false;
        for (int i=head1[u]; i!=-1; i=edges[i].next)
        {
            int v=edges[i].to;
            int w=edges[i].w;
            if (dis1[v]>dis1[u]+w)
            {
                dis1[v]=dis1[u]+w;
                if (!vis[v])
                {
                    vis[v]=true;
                    que.push(v);
                }
            }
        }
    }
}
////========================================
int aa[60080][3],bb[60080][3];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&aa[i][0],&aa[i][1],&aa[i][2]);
            addedge1(aa[i][0],aa[i][1],aa[i][2]);
            addedge1(aa[i][1],aa[i][0],aa[i][2]);
        }
        spfa(1,n);
        memcpy(dis2,dis1,sizeof(dis1));
        //printf("dis2->%d\n",dis2[n]);
        spfa(n,n);
       // printf("dis1->%d\n",dis1[1]);
        int k=0;
        for(int i=0;i<m;i++)
        {
            if(dis2[aa[i][0]]>dis2[aa[i][1]])
                swap(aa[i][0],aa[i][1]);
           // printf("n-%d:%d-1 %d %d %d\n",aa[i][1],aa[i][0],dis1[aa[i][1]],aa[i][2],dis2[aa[i][0]]);
            if(dis1[aa[i][1]]+aa[i][2]+dis2[aa[i][0]]==dis2[n])
            {
                bb[k][0]=aa[i][0];
                bb[k++][1]=aa[i][1];
               // printf("%d %d\n",bb[k-1][0],bb[k-1][1]);
            }
        }
        prepare(n+1,1,n);
        for(int i=0;i<k;i++)
        {
            addedge(bb[i][0],bb[i][1],1);
        }
        int ans1=Dinic_flow();
        init();
        for(int i=0;i<k;i++)
        {
            addedge1(bb[i][0],bb[i][1],1);
            addedge1(bb[i][1],bb[i][0],1);
        }
        spfa(1,n);
        //printf("%d\n",dis2[n]);
        int ans2=m-dis1[n];
        printf("%d %d\n",ans1,ans2);
    }
    return 0;
}

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hdu5294||2015多校联合第一场1007 最短路+最大流

标签：

原文地址：http://blog.csdn.net/lvshubao1314/article/details/46992109