POJ 3189 枚举+最大流判可行性 (二)

ed[num].s = e ;
ed[num].e = s ;
ed[num].l = 0 ;
ed[num].next = head[e] ;
head[e] = num ++ ;
}
int cow[Max][100] ;
int w[Max] ;
int n , m ;
int S, T;
void build(int k,int k1)
{
S = 0 ;
T = n + m + 1 ;
REP(i,1,n)
{
REP(j,k,k1)
add(cow[i][j] ,i + m ,1) ;
add(i + m ,T,1) ;
}
REP(i,1,m)add(S,i,w[i]) ;
}
int deep[1100] ;
int qe[Max * 100] ;
int dinic_bfs()
{
mem(deep,-1) ;
deep[S] = 0 ;
int h = 0 ,t = 0 ;
qe[h ++ ] = S ;
while(h > t)
{
int tt = qe[t ++ ] ;
for (int i = head[tt] ; ~i ; i = ed[i].next )
{
int e = ed[i].e ;
int l = ed[i].l ;
if(l > 0 && deep[e] == -1)
{
deep[e] = deep[tt] + 1 ;
qe[h ++ ] = e ;
}
}
}
return deep[T] != -1 ;
}
int dinic_dfs(int now ,int f)
{
if(now == T) return f ;
int flow = 0 ;
for (int i = head[now] ; ~i ; i = ed[i].next )
{
int e = ed[i].e ;
int l = ed[i].l ;
if(l > 0 && (f - flow) > 0 && deep[e] == deep[now] + 1)
{
int mm = min(l,f - flow) ;
int nn = dinic_dfs(e,mm) ;
flow += nn ;
ed[i].l -= nn ;
ed[i ^ 1].l += nn ;
}
}
if(!flow) deep[now] = -2 ;
return flow ;
}

int dinic()
{
int flow = 0 ;
while(dinic_bfs())
{
flow += dinic_dfs(S,inf) ;
}
return flow ;
}
int solve()
{
int ans = inf ;
REP(i,1,m)
{
REP(j,i,m)
{
init() ;
build(i ,j) ;
int aa = dinic() ;
if(aa == n){
ans = min(ans,j - i + 1) ;
if(ans == 1)
return ans ;
}
}
}
return ans ;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("acm.txt", "r", stdin);
#endif
scanf("%d%d",&n,&m) ;
REP(i,1,n)REP(j,1,m)scanf("%d",&cow[i][j]) ;
REP(i,1,m )scanf("%d",&w[i]) ;
printf("%d\n",solve()) ;
return 0;
}