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python/TangDou/AcWing/BeiBao/734_ErWei.cpp
HuangHai 1f397eca87 'commit'
2025-08-30 18:35:01 +08:00

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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const int M = 10010;
int n;
struct Node {
int s; // 吃掉这块能量石需要花费的时间为s秒
int e; // 获得e个能量
int l; // 不吃的话,每秒失去l个能量
const bool operator<(const Node &b) const {
return s * b.l < b.s * l; // 结构体对比函数
}
} q[N]; // 能量石的数组
int f[N][M];
int idx; // 输出是第几轮的测试数据
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
int m = 0; // 能量视为体积
int res = 0; // 记录最大值
for (int i = 1; i <= n; i++) {
scanf("%d %d %d", &q[i].s, &q[i].e, &q[i].l);
m += q[i].s; // 极限容量,做为背包的上限
}
// 排序
sort(q + 1, q + 1 + n);
// 每次清空状态数组
memset(f, 0, sizeof f);
// 二维01背包
for (int i = 1; i <= n; i++) {
int l = q[i].l, s = q[i].s, e = q[i].e;
for (int j = 0; j <= m; j++) {
f[i][j] = f[i - 1][j]; // 如果不要物品i
if (j >= s) // 如果可以要物品i
f[i][j] = max(f[i - 1][j], f[i - 1][j - s] + max(0, e - (j - s) * l));
res = max(res, f[i][j]);
}
}
cout << "Case #" << ++idx << ": " << res << endl;
}
return 0;
}