[上下界网络流判定] BZOJ 2406 矩阵

上下界网络流：http://www.cnblogs.com/kane0526/archive/2013/04/05/3001108.html

二分答案转化为判定问题：构造矩阵，使得每行每列之和分别满足在一个区间内，这就是带上下界网络流判定问题。

s----------------------->i行----------------------->j列----------------------------->t

     [si-mid,si+mid]                  [L,R]　                [sj-mid,sj+mid]

数据范围看错了 WA了好久...

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

namespace DINIC{
	#define oo 1<<30
	#define V G[p].v
	#define cl(x) memset(x,0,sizeof(x))
	#define M 200000+5
	#define N 1000+5
	struct edge{
		int u,v,f;
		int next;
	};
	edge G[M];
	int head[N],num=1;
	inline void add(int u,int v,int f,int p){
		G[p].u=u; G[p].v=v; G[p].f=f; G[p].next=head[u]; head[u]=p;
	}
	inline void link(int u,int v,int f){
		add(u,v,f,num+=2); add(v,u,0,num^1);
	}
	int S,T;
	int Que[N],l,r;
	int dis[N];
	inline bool bfs(){
		int u;
		memset(dis,-1,sizeof(dis)); cl(Que); l=r=-1;
		dis[S]=1; Que[++r]=S;
		while (l<r)
		{
			u=Que[++l];
			for (int p=head[u];p;p=G[p].next)
				if (G[p].f && dis[V]==-1)
				{
					dis[V]=dis[u]+1;
					Que[++r]=V;
					if (V==T) return true;
				}
		}
		return false;
	} 
	int minimum;
	int cur[N];
	inline bool dfs(int u,int mini){
		if (u==T) { minimum=mini; return true; }
		for (int p=cur[u];p;p=G[p].next)
		{
			cur[u]=p;
			if (G[p].f && dis[V]==dis[u]+1 && dfs(V,min(mini,G[p].f)))
			{
				G[p].f-=minimum; G[p^1].f+=minimum;
				return true;
			}
		}
		dis[u]=-1;
		return false;
	}
	inline int Dinic(){
		int ret=0;
		while (bfs())
		{
			memcpy(cur,head,sizeof(cur));
			while (dfs(S,oo)) 
				ret+=minimum;
		}
		return ret;
	}
	inline void clear(){
		cl(G); cl(head); num=1;
	}
}

int n,m;
int a[205][205];
int sum1[205],sum2[205];
int L,R;
int du[405];

inline bool check(int mid)
{
	using namespace DINIC;
	clear();
	int s=n+m+1,t=n+m+2,u,d;
	S=n+m+3; T=n+m+4;
	memset(du,0,sizeof(du));
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			link(i,j+n,R-L),du[j+n]+=L,du[i]-=L;
	for (int i=1;i<=n;i++)
	{
		d=max(0,sum1[i]-mid),u=sum1[i]+mid;
		link(s,i,u-d);
		du[i]+=d;
		du[s]-=d;
	}
	for (int j=1;j<=m;j++)
	{
		d=max(0,sum2[j]-mid),u=sum2[j]+mid;
		link(j+n,t,u-d);
		du[t]+=d;
		du[j+n]-=d;
	}
	link(t,s,oo);
	int icnt=0,itmp;
	for (int i=1;i<=n+m+2;i++)
		if (du[i]>0)
			link(S,i,du[i]),icnt+=du[i];
		else
			link(i,T,-du[i]);
	itmp=Dinic();
	if (itmp<icnt) return false;
	return true;
}

int main()
{
	int _L,_R,_M;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n); read(m);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			read(a[i][j]),sum1[i]+=a[i][j],sum2[j]+=a[i][j];
	read(L); read(R);
	_L=-1; _R=2000000;
	while (_L+1<_R)
		if (check(_M=(_L+_R)>>1))
			_R=_M;
		else
			_L=_M;
	printf("%d\n",_R);
	return 0;
}