NC25043 Protecting the Flower
Protecting the Flowers
https://ac.nowcoder.com/acm/problem/25043
NC25043 Protecting the Flower
题目地址:
基本思路:
又是这种改排序贪心,我们先只考虑两只牛谁先更优,假如先再那么损耗的花为,假如先再那么损耗的花为,那么我们可以知道只要根据和的大小排序,就能决定这两只牛谁先谁后,那么实际上容易证这个对于更多牛的情况也是普遍成立的,因此根据此规则排序然后直接模拟一遍题意统计答案就行了。
参考代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; #define IO std::ios::sync_with_stdio(false) #define int long long #define rep(i, l, r) for (int i = l; i <= r; i++) #define per(i, l, r) for (int i = l; i >= r; i--) #define mset(s, _) memset(s, _, sizeof(s)) #define pb push_back #define pii pair <int, int> #define mp(a, b) make_pair(a, b) #define INF (int)1e18 inline int read() { int x = 0, neg = 1; char op = getchar(); while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); } while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); } return neg * x; } inline void print(int x) { if (x < 0) { putchar('-'); x = -x; } if (x >= 10) print(x / 10); putchar(x % 10 + '0'); } struct Node{ int t,d; bool operator < (const Node &no) const { return t * no.d < d * no.t;//排序规则; } }; vector<Node> vec; int n,t,d; signed main() { IO; cin >> n; rep(i,1,n){ cin >> t >> d; vec.push_back({t,d}); } sort(vec.begin(),vec.end()); int sum = 0,ans = 0; for(auto it : vec){//直接模拟统计答案; ans += sum * it.d; sum += 2 * it.t; } cout << ans << '\n'; return 0; }