Hdu 1561 The more, The Better
2013-05-01 18:52
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The more, The Better
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3712 Accepted Submission(s): 2186
Problem Description
ACboy很喜欢玩一种战略游戏,在一个地图上,有N座城堡,每座城堡都有一定的宝物,在每次游戏中ACboy允许攻克M个城堡并获得里面的宝物。但由于地理位置原因,有些城堡不能直接攻克,要攻克这些城堡必须先攻克其他某一个特定的城堡。你能帮ACboy算出要获得尽量多的宝物应该攻克哪M个城堡吗?
Input
每个测试实例首先包括2个整数,N,M.(1 <= M <= N <= 200);在接下来的N行里,每行包括2个整数,a,b. 在第 i 行,a 代表要攻克第 i 个城堡必须先攻克第 a 个城堡,如果 a = 0 则代表可以直接攻克第 i 个城堡。b 代表第 i 个城堡的宝物数量, b >= 0。当N = 0, M = 0输入结束。
Output
对于每个测试实例,输出一个整数,代表ACboy攻克M个城堡所获得的最多宝物的数量。
Sample Input
3 2 0 1 0 2 0 3 7 4 2 2 0 1 0 4 2 1 7 1 7 6 2 2 0 0
Sample Output
5 13
Author
8600
Source
HDU 2006-12 Programming Contest
Recommend
LL
思路: 树形DP
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int v = 210; int n, m, f[v]; bool visit[v][v]; int dp[v][v]; void dfs(int index) { int i, j, k; for(i = 1; i <= n; ++i) if(visit[index][i]) { dfs(i); for(j = m; j >= 1; --j) for(k = j - 1; k >= 0; --k) dp[index][j] = max(dp[index][j], dp[index][k] + dp[i][j - k - 1] + f[i]); } } int main() { int i, j, k; while(scanf("%d%d", &n, &m), n || m) { memset(visit, false, sizeof(visit)); memset(dp, 0, sizeof(dp)); int a, b; for(i = 1; i <= n; ++i) { scanf("%d%d", &a, &b); visit[a][i] = true; f[i] = b; } dfs(0); printf("%d\n", dp[0][m]); } }
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