[codeforces551E]GukiZ and GukiZiana

最新推荐文章于 2022-08-29 18:21:59 发布

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最新推荐文章于 2022-08-29 18:21:59 发布

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原文链接：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6803936.html

本文针对CodeForces平台上的551E题目——GukiZ and GukiZiana，提供了一种有效的解决方案。该方案通过分块、排序和二分查找的方式处理数组查询，包括增加指定范围内的元素值以及寻找特定值的最大跨度。

[codeforces551E]GukiZ and GukiZiana

试题描述

Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that a_j = a_i = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:

First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
Second type has form 2 y and asks you to find value of GukiZiana(a, y).

For each query of type 2, print the answer and make GukiZ happy!

输入

The first line contains two integers n, q (1 ≤ n ≤ 5 * 10⁵, 1 ≤ q ≤ 5 * 10⁴), size of array a, and the number of queries.

The second line contains n integers a₁, a₂, ... a_n (1 ≤ a_i ≤ 10⁹), forming an array a.

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 10⁹), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 10⁹), second type query.

输出

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

输入示例

输出示例

数据规模及约定

见“输入”

题解

分块，每块内排序，查找的时候在块内二分。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <cmath>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	int x = 0, f = 1; char c = Getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
	return x * f;
}

#define maxn 500010
#define maxbl 710
#define maxlen 710
#define LL long long

int n, blid[maxn], cb, st[maxbl], en[maxbl], len[maxbl];
LL A[maxn], Ord[maxbl][maxlen], addv[maxbl];

void build_sort(int b) {
	for(int i = st[b]; i <= en[b]; i++) Ord[b][i-st[b]] = A[i];
	sort(Ord[b], Ord[b] + len[b]);
	return ;
}
void pushdown(int b) {
	for(int i = st[b]; i <= en[b]; i++) A[i] += addv[b];
	addv[b] = 0;
	return ;
}

void add(int ql, int qr, int v) {
	if(blid[ql] == blid[qr]) {
		int b = blid[ql];
		pushdown(b);
		for(int i = ql; i <= qr; i++) A[i] += v;
		build_sort(b);
		return ;
	}
	int bl = blid[ql], br = blid[qr];
	pushdown(bl); pushdown(br);
	for(int i = ql; i <= en[bl]; i++) A[i] += v;
	for(int i = st[br]; i <= qr; i++) A[i] += v;
	build_sort(bl); build_sort(br);
	for(int i = bl + 1; i <= br - 1; i++) addv[i] += v;
	return ;
}
bool Find(int b, LL y) {
	int p = lower_bound(Ord[b], Ord[b] + len[b], y - addv[b]) - Ord[b];
	return y - addv[b] == Ord[b][p];
}

int main() {
	n = read(); int q = read();
	int m = sqrt(n + .5);
	for(int i = 1; i <= n; i++) {
		A[i] = read();
		int bl = (i - 1) / m + 1; cb = max(cb, bl);
		blid[i] = bl;
		if(!st[bl]) st[bl] = i; en[bl] = i;
	}
	
	for(int i = 1; i <= cb; i++) len[i] = en[i] - st[i] + 1, build_sort(i);
	while(q--) {
		int tp = read();
		if(tp == 1) {
			int ql = read(), qr = read(), v = read();
			add(ql, qr, v);
		}
		else {
			int y = read(), al = -1, ar = -1;
			for(int i = 1; i <= cb; i++) if(Find(i, y)) {
				pushdown(i); build_sort(i);
				for(int j = st[i]; j <= en[i]; j++) if(y == A[j]) {
					al = j; break;
				}
				break;
			}
			if(al == -1){ puts("-1"); continue; }
			for(int i = cb; i; i--) if(Find(i, y)) {
				pushdown(i); build_sort(i);
				for(int j = en[i]; j >= st[i]; j--) if(y == A[j]) {
					ar = j; break;
				}
				break;
			}
			printf("%d\n", ar - al);
		}
	}
	
	return 0;
}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/6803936.html