学步园

传送门：【SPOJ】839 Optimal Marks

题目分析：第二次看论文终于看懂了，然后去把这题写掉了。

因为标号的每一位都互不影响，所以我们可以单独考虑每一位，对每一位都做一次最小割。对所有已经标号的点，对应位上为1则和源点建边（有向边），否则和汇点建边（有向边），容量都为无穷大，然后所有有边的点都建边（无向边），容量为1，然后跑一遍最小割，之后从源点沿着残余网络dfs一遍，这样被遍历到的点都是标号应该为1的，其余的标号都是0，dfs以后只要在数对应的位置上标上对应的数（0或1）即可。

最小割真是神奇啊~

代码如下：

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

#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i )
#define FOV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )

const int MAXN = 510 ;
const int MAXQ = 510 ;
const int MAXE = 10000 ;
const int INF = 0x3f3f3f3f ;
const int eps = 1e-8 ;

struct Edge {
	int v , c , n ;
	Edge () {}
	Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}
} ;

struct Line {
	int u , v ;
	void input () {
		scanf ( "%d%d" , &u , &v ) ;
	}
} ;

struct NetWork {
	Edge E[MAXE] ;
	int H[MAXN] , cntE ;
	int d[MAXN] , num[MAXN] , pre[MAXN] , cur[MAXN] ;
	int Q[MAXN] , head , tail ;
	int s , t , nv ;
	int flow ;
	int n , m , k ;
	
	Line L[MAXE] ;
	int ans[MAXE] ;
	bool vis[MAXN] ;
	bool mark[MAXN] ;
	
	void init () {
		cntE = 0 ;
		CLR ( H , -1 ) ;
	}
	
	void addedge ( int u , int v , int c , int rc = 0 ) {
		E[cntE] = Edge ( v ,  c , H[u] ) ;
		H[u] = cntE ++ ;
		E[cntE] = Edge ( u , rc , H[v] ) ;
		H[v] = cntE ++ ;
	}
	
	void rev_bfs () {
		CLR ( d , -1 ) ;
		CLR ( num , 0 ) ;
		head = tail = 0 ;
		Q[tail ++] = t ;
		d[t] = 0 ;
		num[d[t]] = 1 ;
		while ( head != tail ) {
			int u = Q[head ++] ;
			for ( int i = H[u] ; ~i ; i = E[i].n ) {
				int v = E[i].v ;
				if ( ~d[v] )
					continue ;
				d[v] = d[u] + 1 ;
				num[d[v]] ++ ;
				Q[tail ++] = v ;
			}
		}
	}
	
	int ISAP () {
		CPY ( cur , H ) ;
		rev_bfs () ;
		flow = 0 ;
		int u = pre[s] = s ;
		while ( d[s] < nv ) {
			if ( u == t ) {
				int f = INF ;
				int pos ;
				for ( int i = s ; i != t ; i = E[cur[i]].v )
					if ( f > E[cur[i]].c ) {
						f = E[cur[i]].c ;
						pos = i ;
					}
				for ( int i = s ; i != t ; i = E[cur[i]].v ) {
					E[cur[i]].c -= f ;
					E[cur[i] ^ 1].c += f ;
				}
				u = pos ;
				flow += f ;
			}
			for ( int &i = cur[u] ; ~i ; i = E[i].n )
				if ( E[i].c && d[u] == d[E[i].v] + 1 )
					break ;
			if ( ~cur[u] ) {
				pre[E[cur[u]].v] = u ;
				u = E[cur[u]].v ;
			}
			else {
				if ( 0 == ( -- num[d[u]] ) )
					break ;
				int mmin = nv ;
				for ( int i = H[u] ; ~i ; i = E[i].n )
					if ( E[i].c && mmin > d[E[i].v] ) {
						mmin = d[E[i].v] ;
						cur[u] = i ;
					}
				d[u] = mmin + 1 ;
				num[d[u]] ++ ;
				u = pre[u] ;
			}
		}
		return flow ;
	}
	
	void dfs ( int u ) {
		vis[u] = 1 ;
		for ( int i = H[u] ; ~i ; i = E[i].n )
			if ( E[i].c && !vis[E[i].v] )
				dfs ( E[i].v ) ;
	}
	
	void solve () {
		int x , y ;
		scanf ( "%d%d" , &n , &m ) ;
		s = 0 ;
		t = n + 1 ;
		nv = t + 1 ;
		CLR ( mark , 0 ) ;
		CLR ( ans , 0 ) ;
		REP ( i , 0 , m )
			L[i].input () ;
		scanf ( "%d" , &k ) ;
		REP ( i , 0 , k ) {
			scanf ( "%d%d" , &x , &y ) ;
			mark[x] = 1 ;
			ans[x] = y ;
		}
		REP ( i , 0 , 31 ) {
			init () ;
			FOR ( u , 1 , n )
				if ( mark[u] ) {
					if ( ans[u] & ( 1 << i ) )
						addedge ( s , u , INF ) ;
					else
						addedge ( u , t , INF ) ;
				}
			REP ( u , 0 , m )
				addedge ( L[u].u , L[u].v , 1 , 1 ) ;
			ISAP () ;
			CLR ( vis , 0 ) ;
			dfs ( s ) ;
			FOR ( u , 1 , n )
				if ( !mark[u] && vis[u] )
					ans[u] |= 1 << i ;
		}
		FOR ( i , 1 , n )
			printf ( "%d\n" , ans[i] ) ;
	}
} x ;

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- )
		x.solve () ;
	return 0 ;
}