[BZOJ2406]矩阵（二分+有源汇有上下界的可行流）

2019-11-14 09:20:10

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题目描述

传送门

题解

刚开始没看见绝对值。。。把这道题翻译一下其实就是构造一个b矩阵，其中每一个点有限制[L,R]，令矩阵c=a-b，使c矩阵每一行的和的绝对值和每一列的和的绝对值的最大值最小

最大值最小很容易想到二分二分答案mid之后，用网络流判定就是满足|∑ai−∑bi|≤mid $|/sum a_i-/sum b_i|/le mid$ 分类讨论一下得出∑ai−mid≤∑bi≤∑ai+mid $/sum a_i-mid/le /sum b_i/le /sum a_i+mid$ 然后就很容易看出是一个有上下界的网络流了原图：每一行和每一列建一个点xi,yi s->xi,[ai-mid,ai+mid] yj->t,[aj-mid,aj+mid] xi->yj,[L,R] 只要判断是否有可行流就行了按照有源汇有上下界的可行流将原图改造求解即可

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>using namespace std;#define N 410#define E 100005#define inf 2000000000int n,m,s,t,ss,tt,L,R,in,out,maxflow,ans;int a[N][N],line[N],lie[N];int tot,point[N],nxt[E],v[E],remain[E];int d[N],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 ans=inf,now=t; 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]; } }}bool check(int mid){ tot=-1;memset(point,-1,sizeof(point)); memset(num,0,sizeof(num)); memset(d,0,sizeof(d)); in=out=0;maxflow=0; for (int i=1;i<=n;++i) { addedge(s,i,mid+mid); d[s]-=line[i]-mid,d[i]+=line[i]-mid; } for (int i=1;i<=m;++i) { addedge(n+i,t,mid+mid); d[n+i]-=lie[i]-mid,d[t]+=lie[i]-mid; } for (int i=1;i<=n;++i) for (int j=1;j<=m;++j) { addedge(i,n+j,R-L); d[i]-=L,d[n+j]+=L; } addedge(t,s,inf); for (int i=1;i<=t;++i) { if (d[i]>0) addedge(ss,i,d[i]),in+=d[i]; if (d[i]<0) addedge(i,tt,-d[i]),out-=d[i]; } if (in!=out) return 0; isap(ss,tt); return maxflow==in;}int find(){ int l=0,r=2000000,mid,ans=2000000; while (l<=r) { mid=(l+r)>>1; if (check(mid)) ans=mid,r=mid-1; else l=mid+1; } return ans;}int main(){ scanf("%d%d",&n,&m); s=n+m+1,t=s+1,ss=t+1,tt=ss+1; for (int i=1;i<=n;++i) for (int j=1;j<=m;++j) { scanf("%d",&a[i][j]); line[i]+=a[i][j]; lie[j]+=a[i][j]; } scanf("%d%d",&L,&R); ans=find(); PRintf("%d/n",ans);}

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