有的堡垒攻克需要攻克另一个堡垒,形成一个森林,最多攻克m个堡垒,求获得宝物的最大价值。
1,以0做根将森林形成树;
2,用背包计算当前节点下需要攻克k个堡垒能获得的宝物最大价值,但是注意同一个根节点的情况不能够先后放入背包,否则会有比如1节点选2个和选三个形成了选5个,也就是某些节点重复计算了。所以要在back第j格时将所有种k依次放入,j--;
#include<iostream> #include<string.h> #include<stdio.h> #include<vector> using namespace std; const int maxa = 205; int dp[maxa][maxa]; int back[maxa][maxa]; int vis[maxa][maxa]; int v[maxa]; vector<int> edge[maxa]; int numb[maxa]; int dfs(int x, int num ){//printf("%d %d\n", x, num); if(vis[x][num] || num == 0) return dp[x][num]; memset(back[x], 0, sizeof(back[x])); for(int i = 0; i < edge[x].size(); i ++){ int k = edge[x][i];//printf("%d ", k); int last = -1; for(int h = num-1; h >= 1; h--){ for(int j = 1; j <= h && j <= numb[k]; j++){ int a = dfs(k, j); back[x][h] = max(back[x][h], back[x][h-j] + a);//printf("%d ", back[h]); }//puts(""); } } vis[x][num] = 1; return dp[x][num] = back[x][num-1]+v[x]; } int dfs1(int x){ int sum = 0; for(int i = 0; i < edge[x].size(); i++){ int k = edge[x][i]; sum += dfs1(k); } return numb[x] = sum +1; } int main(){ int n, m; //freopen("in.cpp", "r", stdin); while(scanf("%d%d", &n, &m), n+m){ memset(vis, 0, sizeof(vis)); for(int i =0; i <= n; i++) edge[i].clear(); for(int i =1; i <= n; i++){ int a, b; scanf("%d%d", &a, &b); v[i] = b; edge[a].push_back(i); } dfs1(0); memset(dp, 0, sizeof(dp)); printf("%d\n", dfs(0, m+1)); /* for(int i = 0; i <= n; i++){ printf("*%d ", numb[i]); }*/ } }
时间: 2024-11-09 01:37:43