[BZOJ2502]清理雪道(有源汇有上下界的最小流)

    xiaoxiao2021-03-26  29

    题目描述

    传送门

    题解

    建原图很简单: 对于能到达的点x,y,x->y,[1,inf] s->i,[0,inf];i->t,[0,inf]

    将原图进行改造 建立附加源汇ss,tt 对于原图中有的边x->y,[b,c],连边x->y,c-b 记某一个点的权d(i)为所有流入这个点的边的下界和-所有流出这个点的边的下界和 若d(i)>0,连边ss->i,d(i) 若d(i)<0,连边i->tt,-d(i) 然后对ss->tt跑最大流 然后连边t->s,inf 再对ss->tt跑最大流 此时t->s,inf这条边的实际流量就是原图中的最小流

    代码

    #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<queue> using namespace std; #define N 110 #define E 30005 #define inf 1000000000 int n,m,x,s,t,ss,tt,maxflow; int tot,point[N],nxt[E],v[E],remain[E]; int d[N],deep[N],num[N],last[N],cur[N]; queue <int> q; void addedge(int x,int y,int cap) { ++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap; ++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0; } void bfs(int t) { for (int i=1;i<=t;++i) deep[i]=t; deep[t]=0; for (int i=1;i<=t;++i) cur[i]=point[i]; while (!q.empty()) q.pop(); q.push(t); while (!q.empty()) { int now=q.front();q.pop(); for (int i=point[now];i!=-1;i=nxt[i]) if (deep[v[i]]==t&&remain[i^1]) { deep[v[i]]=deep[now]+1; q.push(v[i]); } } } int addflow(int s,int t) { int now=t,ans=inf; while (now!=s) { ans=min(ans,remain[last[now]]); now=v[last[now]^1]; } now=t; while (now!=s) { remain[last[now]]-=ans; remain[last[now]^1]+=ans; now=v[last[now]^1]; } return ans; } void isap(int s,int t) { bfs(t); for (int i=1;i<=t;++i) ++num[deep[i]]; int now=s; while (deep[s]<t) { if (now==t) { maxflow+=addflow(s,t); now=s; } bool has_find=false; for (int i=cur[now];i!=-1;i=nxt[i]) if (deep[v[i]]+1==deep[now]&&remain[i]) { has_find=true; cur[now]=i; last[v[i]]=i; now=v[i]; break; } if (!has_find) { int minn=t-1; for (int i=point[now];i!=-1;i=nxt[i]) if (remain[i]) minn=min(minn,deep[v[i]]); if (!(--num[deep[now]])) break; ++num[deep[now]=minn+1]; cur[now]=point[now]; if (now!=s) now=v[last[now]^1]; } } } int main() { tot=-1;memset(point,-1,sizeof(point)); scanf("%d",&n); s=n+1,t=s+1,ss=t+1,tt=ss+1; for (int i=1;i<=n;++i) { scanf("%d",&m); while (m--) { scanf("%d",&x); addedge(i,x,inf); --d[i],++d[x]; } addedge(s,i,inf); addedge(i,t,inf); } for (int i=1;i<=t;++i) { if (d[i]>0) addedge(ss,i,d[i]); if (d[i]<0) addedge(i,tt,-d[i]); } isap(ss,tt); addedge(t,s,inf); isap(ss,tt); printf("%d\n",remain[tot]); }
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