hdu 2883 kebab(存个最大流模板)

来源：互联网发布：国家电网考试软件编辑：程序博客网时间：2024/06/03 20:45
#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>#define ll __int64using namespace std ;const int N = 11111 ;const int M = 2222222 ;const ll INF = (ll) 11111111 * 11111111 ;struct Task {    int l , r , t , id ;} task[N] ;struct Edge {    int to , next ;    ll cap ;} edge[M] ;int head[N] , tot , x[N] ;void new_edge ( int from , int to , ll cap ) { //   printf ( "from = %d , to = %d , cap = %I64d\n" , from , to , cap ) ;    edge[tot].to = to ;    edge[tot].cap = cap ;    edge[tot].next = head[from] ;    head[from] = tot ++ ;    edge[tot].to = from ;    edge[tot].cap = 0 ;    edge[tot].next = head[to] ;    head[to] = tot ++ ;} ;struct Max_Flow {    queue<int> Q ;    ll dis[N] ; int cur[N] ;    int s , t ;    bool bfs ( int n ) {        int u , v , i ;        for ( i = 1 ; i <= n ; i ++ ) dis[i] = INF ;        Q.push ( s ) ; dis[s] = 0 ;        while ( !Q.empty () ) {            u = Q.front () , Q.pop () ;            for ( i = head[u] ; i != -1 ; i = edge[i].next ) {                v = edge[i].to ;                if ( edge[i].cap && dis[v] == INF ) {                    dis[v] = dis[u] + 1 ;                    Q.push ( v ) ;                }            }        }        return dis[t] != INF ;    }    ll dfs ( int u , ll a ) {        if ( u == t || a == 0 ) return a ;        ll flow = 0 , f ;        for ( int& i = cur[u] ; i != -1 ; i = edge[i].next ) {            int v = edge[i].to ;            if ( dis[v] == dis[u] + 1 && (f = dfs (v , min(a,edge[i].cap))) ) {                edge[i].cap -= f ;                edge[i^1].cap += f ;                flow += f ;                a -= f ;                if ( a == 0 ) break ;            }        }        return flow ;    }    ll dinic ( int s , int t , int n ) {        this->s = s ; this->t = t ;        ll ret = 0 ;        while ( bfs ( n ) ){            for ( int i = 1 ; i <= n ; i ++ ) cur[i] = head[i] ;            ret += dfs ( s , INF ) ;        }   //     printf ( "ret = %d\n" , ret ) ;        return ret ;    }} AC ;int cnt ;void init () {    cnt = tot = 0 ;    memset ( head , -1 , sizeof ( head ) ) ;}int main () {    int n , m ;    while ( scanf ( "%d%d" , &n , &m ) != EOF ) {        init () ;        int s = ++ cnt , t = ++ cnt ;        int T = 0 ;        ll need = 0 ;        for ( int i = 1 ; i <= n ; i ++ ) {            int a , b , c , d ;            scanf ( "%d%d%d%d" , &a , &b , &c , &d ) ;            a ++ ;            x[++T] = a , x[++T] = c ;            task[i].l = a , task[i].r = c ;            task[i].t = b * d ;            task[i].id = ++ cnt ;            new_edge ( s , cnt , task[i].t ) ;            need += task[i].t ;        }        sort ( x + 1 , x + T + 1 ) ;        T = unique ( x + 1 , x + T + 1 ) - x - 1 ;        for ( int i = 1 ; i <= T ; i ++ ) {            cnt ++ ;            new_edge ( cnt , t , m ) ;         //   printf ( "cnt = %d , m = %d\n" , cnt , m ) ;            for ( int j = 1 ; j <= n ; j ++ ) {             //   printf ( "l = %d , r = %d , i = %d\n" ) ;                if ( task[j].l <= x[i] && task[j].r >= x[i] ) {                    new_edge ( task[j].id , cnt , INF ) ;               //     printf ( "id = %d\n" , task[j].id ) ;                }            }            if ( i == T || x[i] + 1 >= x[i+1] ) continue ;            cnt ++ ;            new_edge ( cnt , t , m * (x[i+1] - x[i] - 1) ) ;       //     printf ( "cnt = %d , fuck = %d\n" , m * (x[i+1] - x[i] - 1) ) ;            int l = x[i] + 1 , r = x[i+1] - 1 ;            for ( int j = 1 ; j <= n ; j ++ ) {                if ( task[j].l <= l && task[j].r >= r ) {                    new_edge ( task[j].id , cnt , m * ( r - l + 1 ) ) ;                }            }        }        if ( need == AC.dinic ( s , t , cnt ) ) puts ( "Yes" ) ;        else puts ( "No" ) ;    }    return 0 ;}/*1 11 1 4 4No1 21 1 4 4Yes1 11 1 1 14 41 1 6 23 2 8 24 3 7 15 4 10 14 41 1 6 33 2 8 34 3 7 25 4 10 24 31 1 6 33 2 8 34 3 7 25 4 10 2*/
0 0