日常训练2025-1-8
E小红的双生英雄
https://ac.nowcoder.com/acm/contest/99784/E
思路
读题后跟容易发现是一道分组背包的题，转移也比较简单。有一个做动态规划题的技巧是，如果题目相较于传统的DP题有一些其他的约束条件，则把约束条件写成DP的一个维度就行。
代码
#include <bits/stdc++.h>typedef std::pair<long long, long long> pll;
typedef std::pair<int, int> pii;
#define INF 0x3f3f3f3f
#define MOD 998244353
using i64 = long long;
const int N = 1e5+5;void solve(){int n, C, m;std::cin >> n >> C >> m;std::vector<pll> tmp(n+1);std::vector<bool> st(n+1);for (int i = 1; i <= n; i++) {int cost, v;std::cin >> cost >> v;tmp[i] = pll(cost, v);}std::vector<std::vector<pll>> a;for (int i = 0; i < m; i++){int u, v, w;std::cin >> u >> v >> w;st[u] = st[v] = true;a.push_back({tmp[u], tmp[v], {tmp[u].first + tmp[v].first, tmp[u].second + tmp[v].second + w}});}for (int i = 1; i <= n; i++){if (!st[i]) a.push_back({tmp[i]});}std::vector<std::vector<i64>> dp(C+1, std::vector<i64>(5));std::vector<std::vector<i64>> ndp(C+1, std::vector<i64>(5));for (auto s : a){ndp = dp;for (int i = 0; i < s.size(); i++){auto [cost, atk] = s[i];int ok = i == 2;for (int j = C; j >= cost; j--){for (int k = 4; k >= 1 + ok; k--){dp[j][k] = std::max(dp[j][k], ndp[j-cost][k-1-ok] + atk);}}}}i64 ans = 0;for (int j = 0; j <= C; j++){for (int k = 0; k <= 4; k++){ans = std::max(ans, dp[j][k]);}}std::cout << ans << '\n';}signed main()
{std::ios::sync_with_stdio(false);std::cin.tie(nullptr);std::cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);int t = 1, i;for (i = 0; i < t; i++){solve();}return 0;
}
