[BZOJ2406]矩陣（二分+有源匯有上下界的可行流）

2019-11-14 09:20:10

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題目描述

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題解

剛開始沒看見絕對值。。。把這道題翻譯一下其實就是構造一個b矩陣，其中每一個點有限制[L,R]，令矩陣c=a-b，使c矩陣每一行的和的絕對值和每一列的和的絕對值的最大值最小

最大值最小很容易想到二分二分答案mid之后，用網絡流判定就是滿足|∑ai?∑bi|≤mid $|/sum a_i-/sum b_i|/le mid$ 分類討論一下得出∑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個有上下界的網絡流了原圖：每一行和每一列建一個點xi,yi s->xi,[ai-mid,ai+mid] yj->t,[aj-mid,aj+mid] xi->yj,[L,R] 只要判斷是否有可行流就行了按照有源匯有上下界的可行流將原圖改造求解即可

代碼

#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();