看了许多的题解,都有题目翻译,很不错,以后我也这样写。直接翻译样例:
1 1 4 6 /*鞋子的数量N[1, 100]; 拥有的金钱M[1, 1w]; 品牌数目[1, 10]*/ 2 2 5 7 /*以下四行是对于每双鞋的描述*/ 3 3 4 99 /*品牌种类a; 标价b; 高兴程度增加量c*/ 4 1 55 77 5 2 44 66 6 7 /*每一种品牌的鞋子最少买一双,求最大的高兴程度*/
很容易看出是分组背包的题型,trick是价格可能为0(居然有免费的),所以注意dp转移数组初始化-inf。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 #include <cctype> 6 #include <cmath> 7 #include <algorithm> 8 #include <numeric> 9 #include <set> 10 using namespace std; 11 12 //const int = INT_MIN; 13 int s[15][105], v[15][105], dp[15][10005]; 14 15 16 int main () { 17 ios :: sync_with_stdio(false); 18 int N, M, K, a, b, c; 19 while (cin >> N >> M >> K) { 20 int cur[15] = {0}; 21 for (int i = 1; i <= N; ++ i) { 22 cin >> a >> b >> c; 23 s[a][++ cur[a]] = b; 24 v[a][cur[a]] = c; 25 } 26 /*测试种类*/ 27 /* 28 for (int i = 1; i <= N; ++ i) { 29 cout << i << " : " << cur[i] << endl; 30 } 31 */ /*dp数组初始化*/ 32 for (int i = 1; i <= K; ++ i) { 33 for (int j = 0; j <= M; ++ j) { 34 dp[i][j] = INT_MIN; 35 } 36 } 37 38 for (int i = 1; i <= K; ++ i) { 39 for (int j = 1; j <= cur[i]; ++ j) { 40 for (int k = M; k >= s[i][j]; -- k) { 41 dp[i][k] = max (dp[i][k], max (dp[i][k - s[i][j]] + v[i][j], dp[i - 1][k - s[i][j]] + v[i][j]) ); 42 } 43 } 44 } 45 46 //cout << dp[K][M] << endl; 47 if (dp[K][M] < 0) { 48 cout << "Impossible" << endl; 49 } else { 50 cout << dp[K][M] << endl; 51 } 52 } 53 return 0; 54 }
时间: 2024-10-28 20:44:32