bzoj2406 矩阵

我们不妨想一想,这道题目又有\(abs\)又有\(Max\)不是很好算对吧.

所以我们二分答案,考虑怎么\(check\).

对于一个点,显然它能够取的范围是\([l,r]\),接着是对于一行一列都有一个限制使得满足题目条件.

然后直接跑上下界可行流即可.

/*
  mail: mleautomaton@foxmail.com
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=20010,Inf=1e9+10;
int ans,cnt,front[N],dep[N],delta[N],s,t,n,m,a[210][210],l,r,ss,tt,cur[N];queue<int>Q;
int lie[210],hang[210];
struct node{int to,nxt,w;}e[N*100];
void Add(int u,int v,int w){
	e[cnt]=(node){v,front[u],w};front[u]=cnt++;
	e[cnt]=(node){u,front[v],0};front[v]=cnt++;
}
bool bfs(){
	Q.push(ss);memset(dep,0,sizeof(dep));dep[ss]=1;
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=front[u];~i;i=e[i].nxt){
			int v=e[i].to;
			if(!dep[v] && e[i].w){
				dep[v]=dep[u]+1;Q.push(v);
			}
		}
	}
	return dep[tt];
}
int dfs(int u,int flow){
	if(!flow || u==tt)return flow;
	for(int &i=cur[u];~i;i=e[i].nxt){
		int v=e[i].to;
		if(dep[v]==dep[u]+1 && e[i].w){
			int di=dfs(v,min(flow,e[i].w));
			if(di){
				e[i].w-=di;e[i^1].w+=di;return di;
			}
			else dep[v]=0;
		}
	}
	return 0;
}
int Dinic(){
	int flow=0;
	while(bfs()){
		for(int i=0;i<=tt;i++)cur[i]=front[i];
		while(int d=dfs(ss,Inf))flow+=d;
	}
	return flow;
}
int build(int mid){
	memset(front,-1,sizeof(front));cnt=0;int sum=0;
	memset(delta,0,sizeof(delta));
	s=0;t=n+m+1;ss=t+1;tt=ss+1;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			Add(i,j+n,r-l);
			delta[i]-=l;delta[j+n]+=l;
		}
	for(int i=1;i<=n;i++){
		int L=hang[i]-mid,R=hang[i]+mid;
		Add(s,i,R-L);
		delta[s]-=L;delta[i]+=L;
	}
	for(int i=1;i<=m;i++){
		int L=lie[i]-mid,R=lie[i]+mid;
		Add(i+n,t,R-L);
		delta[t]+=L;delta[i+n]-=L;
	}
	for(int i=s;i<=t;i++)
		if(delta[i]>0)Add(ss,i,delta[i]),sum+=delta[i];
		else Add(i,tt,-delta[i]);
	Add(t,s,Inf);
	return sum;
}
bool check(int mid){
	int sum=build(mid);
	int flow=Dinic();
	return flow>=sum;
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in","r",stdin);
#endif
	n=gi();m=gi();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			a[i][j]=gi();
			lie[j]+=a[i][j];hang[i]+=a[i][j];
		}
	l=gi();r=gi();
	int L=0,R=10000000,ret=0;
	while(L<=R){
		int mid=(L+R)>>1;
		if(check(mid)){ret=mid;R=mid-1;}
		else L=mid+1;
	}
	printf("%d\n",ret);
	return 0;
}