Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 23977 Accepted
Submission(s): 9729
//#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<string> #include<algorithm> #define MAX 1100 using namespace std; int dp[MAX][MAX];//行代表个数,列代表体积 int w[MAX], v[MAX], n, W; int _search(int i, int j){ if (dp[i][j] >= 0){ return dp[i][j]; } int res; if (i == n){ res = 0; } else if (w[i] > j){ res = _search(i + 1, j); } else{ res = max(_search(i + 1, j), _search(i + 1, j - w[i]) + v[i]); } return dp[i][j] = res; } int main() { int T; cin >> T; while (T--){ cin >> n >> W; for (int i = 0; i < n; i++){ cin >>v[i]; } for (int i = 0; i < n; i++){ cin >> w[i]; } memset(dp, -1, sizeof(dp)); cout << _search(0,W) << endl; } system("pause"); return 0; }
做这道题的时候二逼了。其实主要是练习记忆搜索。看来递归并没有我想象中掌握的那么好,革命啥时候能成功啊。。。
原文:http://www.cnblogs.com/littlehoom/p/3550547.html