[BZOJ3144][Hnoi2013]切糕（最小割）

2019-11-11 05:24:46

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供稿：網(wǎng)友

題目描述

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題解

很典型的最小割啊切糕的每一小塊就是一個點(diǎn)，然后再在切糕的最上面一層建一層P*Q個虛擬的點(diǎn) s向最上面一層點(diǎn)連邊，最下面一層點(diǎn)向t連邊對于第一個限制，每一個點(diǎn)向它下面的那個點(diǎn)連邊，容量為下面那個點(diǎn)的權(quán)值對于第二個限制，每一個點(diǎn)向它上面第d個點(diǎn)相鄰的四個點(diǎn)連邊，容量為inf，這樣就保證了如果割的不是距離d以內(nèi)的點(diǎn)，源點(diǎn)和匯點(diǎn)還是連通的，一定不是一種最小割的方案然后做最小割就行了

代碼

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>using namespace std;#define N 70005#define E 700005#define inf 2000000000int P,Q,R,d,s,t,maxflow;int val[45][45][45];int tot,point[N],nxt[E],v[E],remain[E];int deep[N],last[N],num[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%d%d",&P,&Q,&R); scanf("%d",&d); s=P*Q*(R+1)+1,t=s+1; for (int k=1;k<=R;++k) for (int i=1;i<=P;++i) for (int j=1;j<=Q;++j) scanf("%d",&val[i][j][k]); for (int k=R+1;k>1;--k) for (int i=1;i<=P;++i) for (int j=1;j<=Q;++j) { int up=(k-1)*P*Q+(i-1)*Q+j; int down=up-P*Q; addedge(up,down,val[i][j][k-1]); } for (int k=1;k<=R;++k) { if (k+d>R) break; for (int i=1;i<=P;++i) for (int j=1;j<=Q;++j) { int down=(k-1)*P*Q+(i-1)*Q+j; int up=down+d*P*Q; if (i!=1) addedge(down,up-Q,inf); if (j!=1) addedge(down,up-1,inf); if (i!=P) addedge(down,up+Q,inf); if (j!=Q) addedge(down,up+1,inf); } } for (int i=1;i<=P;++i) for (int j=1;j<=Q;++j) addedge(s,(i-1)*Q+j+P*Q*R,inf),addedge((i-1)*Q+j,t,inf); isap(s,t);