【BZOJ2333】【SCOI2011】棘手的操作

最新推荐文章于 2022-03-31 20:34:04 发布

cz_xuyixuan

最新推荐文章于 2022-03-31 20:34:04 发布

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分类专栏：【OJ】BZOJ 【类型】做题记录【数据结构】线段树【数据结构】可并堆【算法】离线操作

本文链接：https://blog.youkuaiyun.com/qq_39972971/article/details/79610873

【类型】做题记录同时被 3 个专栏收录

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【OJ】BZOJ

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【数据结构】线段树

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本文介绍了一种使用线段树进行区间更新和查询的方法，适用于动态修改区间内元素值并快速获取最大值的场景。文章通过具体实现细节展示了如何构建线段树、更新区间值及查询区间最大值。

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【题目链接】

点击打开链接

【思路要点】

补档博客，无题解。

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN	300005
#define INF	1e9
struct Node {
	int lc, rc;
	int maxnum, tag;
};
Node a[MAXN * 2];
int n, q, timer, size, dfn[MAXN], home[MAXN];
int f[MAXN], value[MAXN], rit[MAXN];
int x[MAXN], y[MAXN], z[MAXN];
char opt[MAXN][3];
vector <int> t[MAXN];
int F(int x) {
	if (f[x] == x) return x;
	else return f[x] = F(f[x]);
}
void dfs(int pos) {
	dfn[pos] = ++timer;
	for (unsigned i = 0; i < t[pos].size(); i++)
		dfs(t[pos][i]);
}
void update(int x) {
	a[x].maxnum = max(a[a[x].lc].maxnum, a[a[x].rc].maxnum);
}
void pushdown(int x) {
	a[a[x].lc].tag += a[x].tag;
	a[a[x].lc].maxnum += a[x].tag;
	a[a[x].rc].tag += a[x].tag;
	a[a[x].rc].maxnum += a[x].tag;
	a[x].tag = 0;
}
void build(int pos, int l, int r) {
	if (l == r) {
		a[pos].maxnum = value[home[l]];
		return;
	}
	int mid = (l + r) / 2;
	a[pos].lc = ++size;
	build(size, l, mid);
	a[pos].rc = ++size;
	build(size, mid + 1, r);
	update(pos);
}
void maintain(int pos, int l, int r, int ql, int qr, int v) {
	if (l == ql && r == qr) {
		a[pos].tag += v;
		a[pos].maxnum += v;
		return;
	}
	int mid = (l + r) / 2;
	pushdown(pos);
	if (mid >= ql) maintain(a[pos].lc, l, mid, ql, min(mid, qr), v);
	if (mid + 1 <= qr) maintain(a[pos].rc, mid + 1, r, max(mid + 1, ql), qr, v);
	update(pos);
}
int query(int pos, int l, int r, int ql, int qr) {
	if (l == ql && r == qr) return a[pos].maxnum;
	int mid = (l + r) / 2, ans = -INF;
	pushdown(pos);
	if (mid >= ql) ans = max(ans, query(a[pos].lc, l, mid, ql, min(mid, qr)));
	if (mid + 1 <= qr) ans = max(ans, query(a[pos].rc, mid + 1, r, max(mid + 1, ql), qr));
	return ans;
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &value[i]);
		f[i] = i; rit[i] = i;
	}
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		scanf("%s", opt[i]);
		if (opt[i][0] == 'U') {
			scanf("%d%d", &x[i], &y[i]);
			int tx = F(x[i]), ty = F(y[i]);
			if (tx == ty) continue;
			f[ty] = tx; rit[tx] = rit[ty];
			t[tx].push_back(ty);
		}
		if (opt[i][0] == 'A') {
			if (opt[i][1] == '1') scanf("%d%d", &x[i], &y[i]);
			if (opt[i][1] == '2') {
				scanf("%d%d", &x[i], &y[i]);
				x[i] = F(x[i]);
				z[i] = rit[x[i]];
			}
			if (opt[i][1] == '3') scanf("%d", &y[i]);
		}
		if (opt[i][0] == 'F') {
			if (opt[i][1] == '1') scanf("%d", &x[i]);
			if (opt[i][1] == '2') {
				scanf("%d", &x[i]);
				x[i] = F(x[i]);
				z[i] = rit[x[i]];
			}
		}
	}
	for (int i = 1; i <= n; i++)
		if (f[i] == i) dfs(i);
	for (int i = 1; i <= n; i++)
		home[dfn[i]] = i;
	build(0, 1, n);
	for (int i = 1; i <= q; i++) {
		if (opt[i][0] == 'A') {
			if (opt[i][1] == '1') maintain(0, 1, n, dfn[x[i]], dfn[x[i]], y[i]);
			if (opt[i][1] == '2') maintain(0, 1, n, dfn[x[i]], dfn[z[i]], y[i]);
			if (opt[i][1] == '3') maintain(0, 1, n, 1, n, y[i]);
		}
		if (opt[i][0] == 'F') {
			if (opt[i][1] == '1') printf("%d\n", query(0, 1, n, dfn[x[i]], dfn[x[i]]));
			if (opt[i][1] == '2') printf("%d\n", query(0, 1, n, dfn[x[i]], dfn[z[i]]));
			if (opt[i][1] == '3') printf("%d\n", query(0, 1, n, 1, n));
		}
	}
	return 0;
}