CF1734D

5668888888888888

题面

我不想讲了（

思路

Level 1

分史莱姆组来吃！
每次能量够就吃一个史莱姆组！

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Level 2

分组条件：

• 可以吃这个史莱姆组来回血
• 可以吃这个史莱姆组以逃脱

不断左右吃即可，吃到没史莱姆组为止。

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Level 3

在统计组的时候，我们使用类似 two pointer 的解法来统计，同时统计出要吃这个组最少要多少生命。不能吃——NO。吃完——YES。

Level Boss

// 试试就逝世

#include <bits/stdc++.h>
using namespace std;

long long pre[200005];
long long a[200005];

vector<pair<long long, long long> > mleft, mright;

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		mleft.clear();
		mright.clear();
		int n, k;
		scanf("%d %d", &n, &k);
		for (int i = 0; i < n; i++) {
			scanf("%lld", &a[i]);
		}
		long long heal = a[k - 1];
		a[k - 1] = 0;
		for (int i = 0; i < n; i++) {
			pre[i] = a[i] + (i ? pre[i - 1] : 0);
		}
		int l = k - 1, r = k - 1;
		for (int i = k; i < n; i++) {
			if (i == n - 1 || pre[i] >= pre[l]) {
				long long worst = 0, cur = 0;
				for (int j = l + 1; j <= i; j++) {
					cur += a[j];
					worst = min(worst, cur);
				}
				mright.emplace_back(cur, -worst);
				l = i;
			}
		}
		for (int i = k - 2; i >= 0; i--) {
			if (i == 0 || pre[i - 1] <= pre[r - 1]) {
				long long worst = 0, cur = 0;
				for (int j = r - 1; j >= i; j--) {
					cur += a[j];
					worst = min(worst, cur);
				}
				mleft.emplace_back(cur, -worst);
				r = i;
			}
		}
		reverse(mleft.begin(), mleft.end());
		reverse(mright.begin(), mright.end());
		bool flag = 1;
		while (mleft.size() && mright.size()) {
			flag = 0;
			if (mleft.back().second <= heal) {
				heal += mleft.back().first;
				mleft.pop_back();
				flag = 1;
			}
			if (mright.back().second <= heal) {
				heal += mright.back().first;
				mright.pop_back();
				flag = 1;
			}
			if (!flag) {
				break;
			}
		}
		if (flag) {
			puts("YES");
		} else {
			puts("NO");
		}
	}
	return 0;
}