题目链接:https://vjudge.net/problem/HDU-2159
FATE Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17569 Accepted Submission(s): 8250
#include <iostream> #include <cstring> #include <algorithm> using namespace std; int dp[110][110]; //dp[j][u]为最多杀j个怪且忍耐度最多为u时所能获得的最多经验 int main() { int n, m, s, k; while (cin >> n >> m >> k >> s) { int a[110], b[110]; for (int i = 0; i < k; i++)cin >> a[i] >> b[i]; memset(dp, 0, sizeof(dp)); int x; int min = 0x3ffff; bool fp = false; for (int i = 0; i < k; i++) { for (int j = 1; j <= s; j++) //以下这两重循环为二维费用背包的两个限制条件,由于每种怪物有无数种,所以是正序(完全背包) { for (int u = b[i]; u <= m; u++) { //其实将dp数组写成三维的更好理解,dp[i][j][u]杀前i种怪最多杀j个且忍耐度最多为u所能获得的最大经验 dp[j][u] = max(dp[j][u], dp[j - 1][u - b[i]] + a[i]); //其实可以将二维费用的形式和01背包的联系起来,更方便理解和记忆 if (dp[j][u] >= n&& min>u) { min = u; //因为要在保证升级的情况下,使剩下的忍耐度尽可能的大 fp = true; } } } } if (fp)cout << m - min << endl; else cout << "-1" << endl; } return 0; }