hdu 4862 Jump (费用流)

hdu 4862 Jump (费用流)

建图：新建一个temp节点，源点像temp连一条流量为k，费用为0的边，表示最多有k条路径。将原图每个点拆为两个点（i,i'）表示流入该点和流出该点。temp像每个i连边，容量为1，费用为0，表示每个点都可以作为起点。为了保证每个点都被走到，每个的i向i’建边，容量为1，费用为-INF，表示费用非常非常小，它对流非常具有吸引力。每个i’向能去的j建边，容量为1，费用为题意所要求的费用，即i’对应的节点到j对应的节点的距离，相同的话，减去权值，注意要建负边。最后，每个i’向汇点建边，容量为1，费用为0，表示每个点都可以做一条路径的终点。

求解：跑费用流，但与往常的费用流不同的是，我们这里首先是要费用最小，而不是流量最大（因为不一定要把k次都用完）。所以这里判断spfa返回的时候，是看能否让费用继续减小，不能则返回0。因为如果有解的话，第一次spfa肯定就会把所有-INF的边都会被流满（因为它非常具有诱惑力），即所有的点都被走了一遍。最后的答案，先看cost<=n*m*INF? 若果成立，则ans=cost + n * m * INF，否则无解。

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std ;const int N = 2222 ;const int M = 222222 ;const int INF = 11111111 ;struct Edge {    int from , to , next , flow , cap , cost ;    Edge () {}    Edge (int a,int b,int c,int d,int e) {        from = a , to = b , flow = c , cap = d , cost = e ;    }} edge[M] ;int head[N] , tot ;void new_edge ( int from , int to , int flow , int cap , int cost ) {    edge[tot] = Edge ( from , to , flow , cap , cost );    edge[tot].next = head[from] ;    head[from] = tot ++ ;    edge[tot] = Edge ( to , from , 0 , 0 , -cost ) ;    edge[tot].next = head[to] ;    head[to] = tot ++ ;}int vis[N] , dis[N] , pre[N] , add[N] ;queue<int> Q ;int spfa ( int s , int t , int& flow , int& cost , int n ) {    int i , u , v ;    for ( i = 1 ; i <= n ; i ++ ) dis[i] = INF ;    dis[s] = 0 , add[s] = INF ;    Q.push ( s ) ; vis[s] = 1 ;    while ( !Q.empty () ) {        u = Q.front () ; Q.pop () , vis[u] = 0 ;  //      printf ( "fuck head[%d] = %d\n" , u , head[u] ) ;        for ( i = head[u] ; i != -1 ; i = edge[i].next ) {            Edge e = edge[i] ;            v = e.to ;   //         printf ( "v = %d\n" , v ) ;            if ( e.cap > e.flow && dis[v] > dis[u] + edge[i].cost ) {   //             printf ( "u = %d , v = %d\n" , u , v ) ;                dis[v] = dis[u] + edge[i].cost ;                add[v] = min ( add[u] , e.cap - e.flow ) ;                pre[v] = i ;                if ( !vis[v] ) Q.push (v) , vis[v] = 1 ;            }        }    } //   printf ( "dis[%d] = %d\n" , t , dis[t] ) ;    if ( dis[t] >= 0 ) return 0 ;    flow += add[t] ; cost += add[t] * dis[t] ; //   printf ( "flow = %d , cost = %d\n" , flow , cost ) ;    u = t ;    while ( u != s ) {        edge[pre[u]].flow += add[t] ;        edge[pre[u]^1].flow -= add[t] ;        u = edge[pre[u]].from ;    }    return 1 ;}int mincost_maxflow ( int s , int t , int n ) {    int flow = 0 , cost = 0 ;    while ( spfa ( s , t , flow , cost , n ) ) ; //   printf ( "flow = %d\n" , flow ) ;    return cost ;}char mp[33][33] ;int cnt ;int c[33][33][2] ;int change ( int x , int y , int z ) {    if ( !c[x][y][z] ) c[x][y][z] = ++ cnt ; //   printf ( "c[%d][%d][%d] = %d\n" , x , y , z , c[x][y][z] ) ;    return c[x][y][z] ;}int main () {    int n , m , k ;    int cas , ca = 0 ;    scanf ( "%d" , &cas ) ;    while ( cas -- ) {        scanf ( "%d%d%d" , &n , &m , &k ) ;        for ( int i = 1 ; i <= n ; i ++ )            scanf ( "%s" , mp[i]+1 ) ;        cnt = tot = 0 ;        memset ( c , 0 , sizeof (c) ) ;        memset ( head , -1 , sizeof ( head ) ) ;        int s , t , temp ;        s = ++ cnt , t = ++ cnt , temp = ++ cnt ;        new_edge ( s , temp , 0 , k , 0 ) ;        for ( int i = 1 ; i <= n ; i ++ )            for ( int j = 1 ; j <= m ; j ++ ) {                new_edge ( change(i,j,0) , change(i,j,1) , 0 , 1 , -INF ) ;                new_edge ( change(i,j,1) , t , 0 , 1 , 0 ) ;                new_edge ( temp , change(i,j,0) , 0 , 1 , 0 ) ;                for ( int k = j + 1 ; k <= m ; k ++ ) {                    int add = -(k - j - 1) ;                    if ( mp[i][j] == mp[i][k] ) add += mp[i][j] - '0' ;                    new_edge ( change(i,j,1) , change(i,k,0) , 0 , 1 , -add ) ;                }                for ( int k = i + 1 ; k <= n ; k ++ ) {                    int add = -(k - i - 1) ;                    if ( mp[i][j] == mp[k][j] ) add += mp[i][j] - '0' ;                    new_edge ( change(i,j,1) , change(k,j,0) , 0 , 1 , -add ) ;                }            }        int fuck = mincost_maxflow ( s , t , cnt ) ;        printf ( "Case %d : " , ++ ca ) ;        if ( fuck > -INF * n * m ) puts ( "-1" ) ;        else printf ( "%d\n" , -(fuck + n * m * INF ) ) ;    }    return 0 ;}/*1111 1 19*/