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: [email protected]

  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;

}