Counting Haybales（差分+离散化）

题目描述
链接：https://ac.nowcoder.com/acm/contest/7157/H
来源：牛客网

Farmer John has just arranged his N haybales (1≤N≤100,000) at various points along the one-dimensional road running across his farm. To make sure they are spaced out appropriately, please help him answer Q queries (1≤Q≤100,000), each asking for the number of haybales within a specific interval along the road.

输入描述
The first line contains N and Q.
The next line contains N distinct integers, each in the range 0…1,000,000,000, indicating that there is a haybale at each of those locations.
Each of the next Q lines contains two integers A and B (0≤A≤B≤1,000,000,000) giving a query for the number of haybales between A and B, inclusive.

输出描述
You should write Q lines of output. For each query, output the number of haybales in its respective interval.

样例输入
4 6
3 2 7 5
2 3
2 4
2 5
2 7
4 6
8 10
样例输出
2
2
3
4
1
0

题目大意
在区间[0,1000000000]中有一些位置有haybales，求区间[A,B]内haybales的个数

思路
用sum[]数组保存前缀和，sum[i]表示区间[0,i]内haybales的个数，答案为sum[B]-sum[A-1]。但是数组下标无法达到1,000,000,000，我们注意到最多100,000个位置上有haybales，最多100,000个查询，所以实际使用到的下标最多300,000个，我们可以通过离散化处理下标问题。

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5+7;
int mp[N*3], cnt, a[N], x[N], y[N], c[N*3], sum[N*3];
int n, Q;
int main() {
    
    
	freopen("Counting Haybales.txt", "r", stdin);
	scanf("%d%d", &n, &Q);
	for (int i = 1; i <= n; i++) {
    
    
		scanf("%d", &a[i]);
		mp[++cnt] = a[i];
	}
	for (int i = 1; i <= Q; i++) {
    
    
		scanf("%d%d", &x[i], &y[i]);
		mp[++cnt] = x[i];
		mp[++cnt] = y[i];
	}
	sort(mp+1,mp+cnt+1);
	cnt = unique(mp+1,mp+cnt+1)-mp;
	for (int i = 1; i <= n; i++) {
    
    
		int k = lower_bound(mp+1,mp+cnt+1,a[i])-mp;
		c[k]++;
	}
	for (int i = 1; i <= cnt; i++) {
    
    
		sum[i] = sum[i-1] + c[i];
	}
	for (int i = 1; i <= Q; i++) {
    
    
		int xx = lower_bound(mp+1,mp+cnt+1,x[i])-mp;
		int yy = lower_bound(mp+1,mp+cnt+1,y[i])-mp;
		printf("%d\n", sum[yy]-sum[xx-1]);
	}
	return 0;
}

需要注意的点

排序后，使用unique()函数去重
lower_bound()函数返回第一个大于等于的位置，upper_bound()函数返回第一个大于的位置，所以这里使用lower_bound()查找离散化后的下标

Counting Haybales（差分+离散化）

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