【codevs1034】星际转移问题（家园）（网络流）

来源：互联网发布：淘宝html代码生成器编辑：程序博客网时间：2024/04/30 20:04

题目：
我是超链接
题解：
建图如下
这里写图片描述
枚举时间点，因为飞船在跑，由st[i][t]到st[next[i]][t+1]连边
人们也可以蹲在太空站里，st[i][t]到st[i][t+1]连边
然后就是最大流啦
代码：

#include <cstdio>#include <cstring>#include <queue>#include <iostream>#define INF 1e9using namespace std;int p[55],tim,deep[50005],point[50005],cur[50005],tot=-1,r[55][55],next[50005],v[50005],remind[50005];int n,m,k;void addline(int x,int y,int cap){    ++tot; next[tot]=point[x]; point[x]=tot; v[tot]=y; remind[tot]=cap;    ++tot; next[tot]=point[y]; point[y]=tot; v[tot]=x; remind[tot]=0;}int addflow(int t,int now,int limit){    if (!limit||now==t) return limit;    int f=0,flow=0;    for (int i=cur[now];i!=-1;i=next[i])    {        cur[now]=i;        if (deep[v[i]]==deep[now]+1 && (f=addflow(t,v[i],min(remind[i],limit))))        {            flow+=f;            limit-=f;            remind[i]-=f;            remind[i^1]+=f;            if (!limit) break;        }    }    return flow;        }bool bfs(int s,int t){    queue <int> q;    q.push(s);    memset(deep,0x7f,sizeof(deep));    deep[s]=0;    while (!q.empty())    {        int now=q.front();q.pop();        for (int i=point[now];i!=-1;i=next[i])          if (remind[i] && deep[v[i]]>INF)          {            deep[v[i]]=deep[now]+1;            q.push(v[i]);          }    }    if (deep[t]<INF) for (int i=1;i<=m*tim;i++) cur[i]=point[i];    cur[s]=point[s]; cur[t]=point[t];    return deep[t]<INF;}int main(){    int i,j;    memset(point,-1,sizeof(point));    memset(next,-1,sizeof(next));    scanf("%d%d%d",&n,&m,&k);    for (i=1;i<=m;i++)    {        scanf("%d%d",&p[i],&r[i][0]);        for (j=1;j<=r[i][0];j++)        {            scanf("%d",&r[i][j]);            if (r[i][j]<=0) r[i][j]+=5000;                  }    }    int s=5000,t=4999,we=0;    tim=0;      while ((++tim)<=600)    {        for (i=1;i<=m;i++)        {            int x,y;            if (r[i][(tim-1)%r[i][0]+1]==s) x=s;            else if (r[i][(tim-1)%r[i][0]+1]==t) x=t;            else x=(tim-1)*n+r[i][(tim-1)%r[i][0]+1];            if (r[i][tim%r[i][0]+1]==s) y=s;            else if (r[i][tim%r[i][0]+1]==t) y=t;            else  y=tim*n+r[i][tim%r[i][0]+1];            addline(x,y,p[i]);        }        for (i=1;i<=n;i++) addline((tim-1)*n+i,(tim-1)*n+i+n,INF);        while (bfs(s,t)) we+=addflow(t,s,INF);        if (we>=k){printf("%d",tim); return 0;}    }    printf("0");}

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