【POJ】1698 Alice's Chance 最大流
2014-08-03 14:27
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传送门:【POJ】1698 Alice's Chance
题目分析:没想出二分匹配的方法。。偷个懒。。最大流求解,最大流的话就是模板题了。。。
代码如下:
题目分析:没想出二分匹配的方法。。偷个懒。。最大流求解,最大流的话就是模板题了。。。
代码如下:
#include <cmath> #include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define REP( i , a , b ) for ( int i = a ; i < b ; ++ i ) #define REV( i , a , b ) for ( int i = a - 1 ; i >= b ; -- i ) #define FOR( i , a , b ) for ( int i = a ; 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 ) typedef long long LL ; typedef int type_c ; typedef LL type_f ; const int MAXN = 405 ; const int MAXQ = 405 ; const int MAXE = 150000 ; const int INF = 0x3f3f3f3f ; struct Edge { int v , n ; type_c c , rc ; Edge () {} Edge ( int v , type_c c , int n ) : v ( v ) , c ( c ) , n ( n ) {} } ; struct Net { Edge E[MAXE] ; int H[MAXN] , cntE ; int d[MAXN] , cur[MAXN] , pre[MAXN] , num[MAXN] ; int Q[MAXQ] , head , tail ; int s , t , nv ; type_f flow ; int n , m ; bool vis[MAXN] ; void init () { cntE = 0 ; CLR ( H , -1 ) ; } void addedge ( int u , int v , type_c c , type_c 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] == -1 ) { Q[tail ++] = v ; d[v] = d[u] + 1 ; num[d[v]] ++ ; } } } } type_f ISAP () { CPY ( cur , H ) ; rev_bfs () ; flow = 0 ; int u = pre[s] = s , i , pos , mmin ; while ( d[s] < nv ) { if ( u == t ) { type_f f = INF ; for ( i = s ; i != t ; i = E[cur[i]].v ) if ( f > E[cur[i]].c ) { f = E[cur[i]].c ; pos = i ; } for ( 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 ( i = cur[u] ; ~i ; i = E[i].n ) if ( E[i].c && d[u] == d[E[i].v] + 1 ) break ; if ( ~i ) { cur[u] = i ; pre[E[i].v] = u ; u = E[i].v ; } else { if ( 0 == -- num[d[u]] ) break ; for ( mmin = nv , 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 ( !vis[E[i].v] && E[i].c ) dfs ( E[i].v ) ; } void solve () { int day[7] , D , W , sum = 0 ; init () ; s = 400 , t = s + 1 , nv = t + 1 ; scanf ( "%d" , &n ) ; REP ( i , 0 , n ) { REP ( j , 0 , 7 ) scanf ( "%d" , &day[j] ) ; scanf ( "%d%d" , &D , &W ) ; addedge ( s , i + 350 , D ) ; sum += D ; REP ( j , 0 , W ) REP ( k , 0 , 7 ) if ( day[k] ) addedge ( i + 350 , j * 7 + k , 1 ) ; } REP ( i , 0 , 350 ) addedge ( i , t , 1 ) ; ISAP () ; if ( flow == sum ) printf ( "Yes\n" ) ; else printf ( "No\n" ) ; } } e ; int main () { int T ; scanf ( "%d" , &T ) ; while ( T -- ) e.solve () ; return 0 ; }
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