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++;}elser++;}printf( "%d\n",range );}return 0;
}