【hoj2634】【最小割】How to earn more

来源：互联网发布：配电箱大小计算软件编辑：程序博客网时间：2024/06/06 10:07

借用Edelweiss的话，此题是典型的“蕴含式最大获利问题”，使用解决最大权闭合子图的建模方法即可解决。

每个项目i作为一个点并连边(s,i,Ai),每名员工j作为一个点并连边(j,t,Bj),若项目i需要雇佣员工j则连边(i,j,∞)。

设最小割为ans，则∑Ai - ans即为结果

代码:

#include<cstdio>#include<cstring>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 500 + 10;int dis[maxn],gap[maxn],pre[maxn],cur[maxn];int cap[maxn][maxn];int s,t,nodenum;int T,n,m;void init(){freopen("hoj2634.in","r",stdin);freopen("hoj2634.out","w",stdout);}inline int min(int a,int b){return a < b ? a : b;}int sap(){memset(dis,0,sizeof(dis));memset(gap,0,sizeof(gap));memset(pre,0,sizeof(pre));int u = pre[s] = s,maxflow = 0,aug = inf;gap[0] = nodenum;while(dis[s] < nodenum){loop:   for(int v = cur[u];v < nodenum;v++){if(cap[u][v] && dis[u] == dis[v] + 1){pre[v] = u;aug = min(aug,cap[u][v]);u = v;if(v == t){maxflow += aug;for(u = pre[u];v != s;v = u,u = pre[u]){cap[u][v] -= aug;cap[v][u] += aug;}aug = inf;}goto loop;}}int mind = nodenum;for(int v = 0;v < nodenum;v++){if(cap[u][v] && (mind > dis[v])){cur[u] = v;mind = dis[v];}}if((--gap[dis[u]]) == 0)break;gap[dis[u] = mind + 1]++;u = pre[u];}return maxflow;}void readdata(){scanf("%d",&T);while(T--){memset(cap,0,sizeof(cap));scanf("%d%d",&m,&n);int sum = 0;s = 0,t = n + m + 1;nodenum = t + 1;for(int i = 1;i <= m;i++){int tmp;scanf("%d",&tmp);sum += tmp;cap[s][i] = tmp;}for(int i = 1;i <= n;i++){int tmp;scanf("%d",&tmp);cap[i + m][t] = tmp;}for(int i = 1;i <= m;i++){ int tot; scanf("%d",&tot); for(int j = 1;j <= tot;j++) { int tmp; scanf("%d",&tmp); tmp++; cap[i][tmp + m] = inf; }}int ans = sum - sap();printf("%d\n",ans);}}int main(){init();readdata();return 0;}