给了一些任务,然后给了一些完成某些任务的限制,然后又给了限制之间的拓扑关系,最后问你最大收益。
思路:
很直白,就是流的一个应用,最大权闭包,没涉及到什么想法的地方,建图也不坑,直接说建图吧,
s - 所有任务 流量是 任务价值
所有限制 - t 流量是 限制代价
a -> b 流量 INF a限制的拓扑关系在b的后面
最后答案是 所有任务的价值 - maxflow
#include<stdio.h> #include<string.h> #include<queue> #define N_node 100 #define N_edge 8000 #define INF 1000000000 using namespace std; typedef struct { int to ,next ,cost; }STAR; typedef struct { int x ,t; }DEP; STAR E[N_edge]; DEP xin ,tou; int list[N_node] ,listt[N_node] ,tot; int deep[N_node]; void add(int a ,int b ,int c) { E[++tot].to = b; E[tot].cost = c; E[tot].next = list[a]; list[a] = tot; E[++tot].to = a; E[tot].cost = 0; E[tot].next = list[b]; list[b] = tot; } bool BFS_Deep(int s ,int t ,int n) { memset(deep ,255 ,sizeof(deep)); deep[s] = 0; xin.x = s ,xin.t = 0; queue<DEP>q; q.push(xin); while(!q.empty()) { tou = q.front(); q.pop(); for(int k = list[tou.x] ;k ;k = E[k].next) { xin.x = E[k].to; xin.t = tou.t + 1; if(deep[xin.x] != -1 || !E[k].cost) continue; deep[xin.x] = xin.t; q.push(xin); } } for(int i = 0 ;i <= n ;i ++) listt[i] = list[i]; return deep[t] != -1; } int minn(int x ,int y) { return x < y ? x : y; } int DFS_Flow(int s ,int t ,int flow) { if(s == t) return flow; int nowflow = 0; for(int k = listt[s] ;k ;k = E[k].next) { listt[s] = k; int to = E[k].to; int c = E[k].cost; if(deep[to] != deep[s] + 1 || !c) continue; int tmp = DFS_Flow(to ,t ,minn(c ,flow - nowflow)); nowflow += tmp; E[k].cost -= tmp; E[k^1].cost += tmp; if(nowflow == flow) break; } if(!nowflow) deep[s] = 0; return nowflow; } int DINIC(int s ,int t ,int n) { int ans = 0; while(BFS_Deep(s ,t ,n)) { ans += DFS_Flow(s ,t ,INF); } return ans; } int main () { int i ,j ,n ,nn ,m ,a ,T ,cas = 1; int s ,t ,sum_z; scanf("%d" ,&T); while(T--) { scanf("%d %d" ,&n ,&m); s = 0 ,t = n + m + 1; memset(list ,0 ,sizeof(list)) ,tot = 1; for(sum_z = 0 ,i = 1 ;i <= n ;i ++) { scanf("%d" ,&a); sum_z += a; add(s ,i ,a); } for(i = 1 ;i <= m ;i ++) { scanf("%d" ,&a); add(i + n ,t ,a); } for(i = 1 ;i <= n ;i ++) { scanf("%d" ,&nn); while(nn--) { scanf("%d" ,&a); a ++; add(i ,a + n ,INF); } } for(i = 1 ;i <= m ;i ++) for(j = 1 ;j <= m ;j ++) { scanf("%d" ,&a); if(a) add(i + n ,j + n ,INF); } printf("Case #%d: " ,cas ++); printf("%d " ,sum_z - DINIC(s ,t ,t)); } return 0; }