[BZOJ2406]矩阵（二分+有源汇有上下界的可行流）

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 分类讨论一下得出 ∑ai−mid≤∑bi≤∑ai+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 2000000000 int 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); }