Secret Milking Machine POJ

来源：互联网发布：网络用语吃鸡什么意思编辑：程序博客网时间：2024/05/24 07:28

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二分最大边记为lim 不超过lim的边容量记为1 否则记为0 再抽象一个源点从源点到1的容量为题目所给的t 然后以此建图看是否满流

感觉网络流的抽象建图很关键这道题看了别人的建图才会的...

符合条件就是1 即存在反之就是0 即不存在这样就把一个记录路径长度的图转化为记录路径是否可行的图每增广一次图中就少一条路长知识了!

#include <stdio.h>#include <queue>#include <cstring>#include <algorithm>using namespace std;#define N 1e9struct node1{    int u;    int v;    int w;};struct node2{    int v;    int w;    int next;};node1 pree[80010];node2 edge[160010];int first[210],dis[210],gap[210],cur[210],pre[210];int n,m,t,num,sum,ans,ss,ee;void binsearch(int l,int r);void build(int lim);void addedge(int u,int v,int w);void isap();void bfs();int main(){    int i,u,v,w,minn,maxx;    while(scanf("%d%d%d",&n,&m,&t)!=EOF)    {        num=0,minn=N,maxx=0;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            num++;            pree[num].u=u;            pree[num].v=v;            pree[num].w=w;            num++;            pree[num].u=v;            pree[num].v=u;            pree[num].w=w;            if(w<minn) minn=w;            if(w>maxx) maxx=w;        }        binsearch(minn,maxx);        printf("%d\n",ans);    }    return 0;}void binsearch(int l,int r){    int lim;    ans=r;    while(l<=r)    {        lim=(l+r)/2;        build(lim);        isap();        if(sum>=t)        {            r=lim-1;            ans=lim;        }        else        {            l=lim+1;        }    }    return;}void build(int lim){    int i;    memset(first,-1,sizeof(first));    num=0,ss=n+1,ee=n;    addedge(ss,1,t);    addedge(1,ss,0);    for(i=1;i<=m*2;i++)    {        if(pree[i].w<=lim)        {            addedge(pree[i].u,pree[i].v,1);            addedge(pree[i].v,pree[i].u,0);        }        else        {            addedge(pree[i].u,pree[i].v,0);        }    }    return;}void addedge(int u,int v,int w){    edge[num].v=v;    edge[num].w=w;    edge[num].next=first[u];    first[u]=num++;    return;}void isap(){    int j,u,v,flow,minn;    memcpy(cur,first,sizeof(first));    memset(pre,-1,sizeof(pre));    bfs();    num=n+1,sum=0,u=ss,flow=N;    while(dis[ss]<num)    {        int &i=cur[u];        for(;i!=-1;i=edge[i].next)        {            v=edge[i].v;            if(edge[i].w>0&&dis[v]+1==dis[u])            {                pre[v]=i;                u=v,flow=min(flow,edge[i].w);                if(u==ee)                {                    while(u!=ss)                    {                        j=pre[u];                        edge[j].w-=flow;                        edge[j^1].w+=flow;                        u=edge[j^1].v;                    }                    sum+=flow,flow=N;                }                break;            }        }        if(i==-1)        {            if(--gap[dis[u]]==0) break;            cur[u]=first[u];            minn=num-1;            for(j=first[u];j!=-1;j=edge[j].next)            {                if(edge[j].w>0)                {                    minn=min(minn,dis[edge[j].v]);                }            }            dis[u]=minn+1;            gap[dis[u]]++;            if(u!=ss)            {                u=edge[pre[u]^1].v;            }        }    }    return;}void bfs(){    queue <int> que;    int i,u,v;    memset(dis,-1,sizeof(dis));    memset(gap,0,sizeof(gap));    que.push(ee);    dis[ee]=0;    while(!que.empty())    {        u=que.front();        que.pop();        gap[dis[u]]++;        for(i=first[u];i!=-1;i=edge[i].next)        {            v=edge[i].v;            if(dis[v]==-1)            {                dis[v]=dis[u]+1;                que.push(v);            }        }    }    return;}

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