洛谷 P3130 [USACO15DEC] Counting Haybale P

原题链接

题目本质：线段树

感觉我对线段树稍有敏感，线段树一眼就看出来了，思路出来得也快，这道题也并不是很难。

解题思路：

这道题能看出来是线段树就基本成功一半了，区间修改+区间查询，就基本上是裸的线段树，但是用朴素的线段树会超时，得加上懒标记。

代码如下：

//这狗屎出题人第一个就整个线段树
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200010;
ll sum[N << 2];
int mm[N << 2];
ll lazy[N << 2];
int a[N];
int n, m;
char c;
int le, ri, p;
void pushup(int rt) {
	sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
	mm[rt] = min(mm[rt << 1], mm[rt << 1 | 1]);
}
void pushdown(int rt, int l, int r) {
	lazy[rt << 1] += lazy[rt], lazy[rt << 1 | 1] += lazy[rt];
	int mid = (l + r) / 2;
	sum[rt << 1] += lazy[rt] * (mid - l + 1);
	sum[rt << 1 | 1] += lazy[rt] * (r - mid);
	mm[rt << 1] += lazy[rt];
	mm[rt << 1 | 1] += lazy[rt];
	lazy[rt] = 0;
}
void build(int l, int r, int cur) {
	if (l == r) {
		sum[cur] = a[l];
		mm[cur] = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(l, mid, cur << 1);
	build(mid + 1, r, cur << 1 | 1);
	pushup(cur);
}
ll Sum(int L, int R, int l, int r, int cur) {
	if (l >= L && r <= R)
		return sum[cur];
	if (lazy[cur])
		pushdown(cur, l, r);
	int mid = (l + r) >> 1;
	ll res = 0;
	if (mid >= L)
		res += Sum(L, R, l, mid, cur << 1);
	if (mid < R)
		res += Sum(L, R, mid + 1, r, cur << 1 | 1);
	return res;
}
ll Min(int L, int R, int l, int r, int cur) {
	if (l >= L && r <= R)
		return mm[cur];
	if (lazy[cur])
		pushdown(cur, l, r);
	int mid = (l + r) >> 1;
	if (mid >= R)
		return Min(L, R, l, mid, cur << 1);
	else if (mid < L)
		return Min(L, R, mid + 1, r, cur << 1 | 1);
	return min(Min(L, R, l, mid, cur << 1), Min(L, R, mid + 1, r, cur << 1 | 1));
}
void update(int L, int R, int l, int r, int cur, int z) {
	if (l >= L && r <= R) {
		sum[cur] += (ll)z * (r - l + 1);
		mm[cur] += z;
		lazy[cur] += z;
		return;
	}
	if (lazy[cur])
		pushdown(cur, l, r);
	int mid = (l + r) >> 1;
	if (mid >= L)
		update(L, R, l, mid, cur << 1, z);
	if (mid < R)
		update(L, R, mid + 1, r, cur << 1 | 1, z);
	pushup(cur);
}
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	build(1, n, 1);
	for (int i = 0; i < m; i++) {
		cin >> c;
		if (c == 'M') {
			cin >> le >> ri;
			cout << Min(le, ri, 1, n, 1) << '\n';
		} else if (c == 'S') {
			cin >> le >> ri;
			cout << Sum(le, ri, 1, n, 1) << '\n';
		} else {
			cin >> le >> ri >> p;
			update(le, ri, 1, n, 1, p);
		}
	}
	return 0;
}

不对不对，忘了吐嘈出题人了，T1就来一道大线段树，我***