UVALIVE 4819 最大流

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题意:有N场比赛,每场比赛需要一定数量的题目数,现在有M个题目,每个题目只能提供给特定的几场比赛,并且一次只能在一场比赛中出现。

问最多可以举办多少场比赛。

思路:因为N = 15 , 所以直接二进制枚举举办的比赛的情况,然后对于每种情况建图,

S - >题目,流量1

题目 ->比赛,流量1

比赛->T,流量为该场比赛需要的题目数。

每次都跑最大流,看是否等于所需的题目数,然后更新答案即可。

#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#define Max 2505#define FI first#define SE second#define ll long long#define PI acos(-1.0)#define inf 0x3fffffff#define LL(x) ( x << 1 )#define bug puts("here")#define PII pair<int,int>#define RR(x) ( x << 1 | 1 )#define mp(a,b) make_pair(a,b)#define mem(a,b) memset(a,b,sizeof(a))#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )using namespace std;#define N 222#define M 5555struct kdq {    int e , next , l ;} ed[M] ;int head[N] , num ;void init() {    mem(head ,-1) ;    num = 0 ;}void add(int s ,int e ,int l) {    ed[num].e = e ;    ed[num].next = head[s] ;    ed[num].l = l ;    head[s] = num ++ ;    ed[num].e = s ;    ed[num].next = head[e] ;    ed[num].l = 0 ;    head[e] = num ++ ;}map<string ,int>MM;int n , m ;int a[N] ;string x ;int fk[111][111] ;char now[M] ;int S , T ;int deep[111] ;int qe[111111] ;int dinic_bfs() {    mem(deep , -1) ;    deep[S] = 0 ;    int h = 0 , t = 0 ;    qe[h ++ ] = S ;    while(h > t) {        int tp = qe[t ++ ] ;        for (int i = head[tp] ; ~i ; i = ed[i].next ) {            int e = ed[i].e ;            int l = ed[i].l ;            if(l > 0 && deep[e] == -1) {                deep[e] = deep[tp] + 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(deep[e] == deep[now] + 1 && l > 0 && (f - flow) > 0) {            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 main() {    int ca = 0 ;    while(scanf("%d%d",&n,&m) , (n + m)) {        S = 0 , T = n + m + 1 ;        init() ;        MM.clear() ;        mem(fk ,0) ;        for (int i = 1 ; i <= n ; i ++ ) {            cin >> x ;            MM[x] = i ;            scanf("%d",&a[i]) ;        }        string st ;        st.clear() ;        gets(now) ;        for (int i = 1 ; i <= m ; i ++ ) {            gets(now) ;            st.clear() ;            int l = strlen(now) ;            if(l == 0)continue ;            for (int j = 0 ; j < l  ; j ++ ) {                if(now[j] == ' ') {                    if(st.size() == 0)continue ;                    fk[MM[st]][i] = 1 ;                    st.clear() ;                    continue ;                }                st += now[j] ;            }            if(st.size()) {                fk[MM[st]][i] = 1 ;            }        }        int ans = 0 ;        for (int i = 0 ; i < (1 << n) ; i ++ ) {            init() ;            int nk = 0 ;            int sum = 0 ;            for (int j = 1 ; j <= m ; j ++ )add(S , j , 1) ;            for (int j = 0 ; j < n ; j ++ ) {                if(i & (1 << j)) {                    for (int k = 1 ; k <= m ; k ++ ) {                        if(fk[j + 1][k]) {                            add(k , j + 1 + m , 1) ;                        }                    }                    nk ++ ;                    sum += a[j + 1] ;                    add(j + 1 + m , T , a[j + 1]) ;                }            }            int fff = dinic() ;            if(fff == sum)ans = max(ans , nk) ;        }        printf("Case #%d: %d\n",++ ca ,ans) ;    }    return 0 ;}