POJ 3189 图论技巧题

这题的技巧就是：枚举---旋转卡壳法。

题目大意：

每头牛对每个谷仓有一个喜欢程度，FJ的目的就是要使得每头牛的happy值尽可能的相同，求最小的范围。

这题的枚举还是很有技巧的。虽然知道怎么来滑动窗口。。 但是我的网络流EK算法太不给力了！

没办法.. 去学习了下SAP勉强把这题切掉了= =

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#define MN 1111
#define INF 0x0FFFFFFF
#define CC(m,what)		memset(m,what,sizeof(m))
#define FOR(i,a,b)		for( int i = (a) ; i < (b) ; i ++ )
#define FF(i,a)			for( int i = 0 ; i < (a) ; i ++ )
#define FFD(i,a)		for( int i = (a)-1 ; i >= 0 ; i --)
#define SS(a)			scanf("%d",&a)
#define LL(a)			((a)<<1)
#define RR(a)			(((a)<<1)+1)
#define SZ(a)			((int)a.size())
#define PP(n,m,a)		puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
 
#define read			freopen("in.txt","r",stdin)
#define write			freopen("out.txt","w",stdout)
 
#define two(x)			((LL)1<<(x))
#define include(a,b)		(((a)&(b))==(b))
template<class T> inline T countbit(T n)	{return n?1+countbit(n&(n-1)):0;}
template<class T> inline T sqr(T a)	{return a*a;}
template<class T> inline void checkmin(T &a,T b)	{if(a == -1 || a > b)a = b;}
template<class T> inline void checkmax(T &a,T b)	{if(a < b)	a = b;}
using namespace std;

struct EDGE
{	int u,v,len;
}edge[1000*21];

int map[MN][MN],vis[MN],barn[MN],pre[MN];
int N,B,s,t,ans;

bool cmp( EDGE a,EDGE b ){ return a.len<b.len; }

void setG()
{
 	 s=0;t=B+N+1;
 	 int i,j,num,at;
 	 for( i=1;i<=N;i++ )
 	 for( j=1;j<=B;j++ )
 	 {
 	 	  scanf( "%d",&num );
 	 	  at=(i-1)*B+j-1;
 	 	  edge[at].u=num;
 	 	  edge[at].v=B+i;
 	 	  edge[at].len=j;
	 }
	 sort( edge,edge+B*N,cmp );
	 for( i=1;i<=B;i++ )
	 	  scanf( "%d",&barn[i] );
}

void initG(int l,int r)
{
 	 int i,j;
 	 for( i=0;i<=t;i++ )
 	 for( j=0;j<=t;j++ ) map[i][j]=0;
 	 for( i=(l-1)*N;i<r*N;i++ )
 	 	  map[edge[i].u][edge[i].v]++;
	 for( i=1;i<=B;i++ )
	 	  map[s][i]=barn[i];
	 for( i=B+1;i<=B+N;i++ )
	 	  map[i][t]++;
}

int sap() {
	int cur[MN],dis[MN],gap[MN];
	CC(cur,0);CC(dis,0);CC(gap,0);
	int u=pre[s]=s,maxflow=0,aug=-1;
	gap[0]=t+1;
	while( dis[s]<=t ){
loop:
	 	   for( int v=cur[u];v<=t;v++ )
		if( map[u][v]&&dis[u]==dis[v]+1 )
		{
			cur[u]=v;
			checkmin(aug,map[u][v]);
			pre[v]=u;
			u=v;
			if( v==t )
			{
				maxflow+=aug;
				for( u=pre[u];v!=s;v=u,u=pre[u] )
				{
					 map[u][v]-=aug;
					 map[v][u]+=aug;
			    }
				aug=-1;
			}
			goto loop;
		}
		int mind=t;
		for( int v=0;v<=t;v++ )
		if( map[u][v]&&mind>dis[v] )
		{
			cur[u]=v;
			mind=dis[v];
		}
		if( (--gap[dis[u]])==0 ) break;
		gap[dis[u]=mind+1]++;
		u=pre[u];
	}
	return maxflow;
}

bool work()
{
 	 int i,j;
	 if( sap()==N ) return true;
	 else return false;
}

int main()
{
 	while( scanf("%d%d",&N,&B)!=EOF )
 	{
	 	   setG();
	 	   int range=INF;
	 	   int r=1,l=1;
	 	   while( l<=B&&l<=r&&r<=B )
	 	   {
		   		initG(l,r);
		   		if( work() )
		   		{
				 	range=min( range,r-l+1 );
		   			l++;
		        }
	   			else
	   				r++;
	   	   }
		   printf( "%d\n",range );
  	}
  	return 0;
}