洛谷 P3184 [USACO16DEC]Counting Haybales数草垛

题目描述

Farmer John has just arranged his NN haybales (1 \leq N \leq 100,0001≤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 QQ queries (1 \leq Q \leq 100,0001≤Q≤100,000 ), each asking for the number of haybales within a specific interval along the road.

农夫John在一条穿过他的农场的路上（直线）放了N个干草垛（1<=N<=100,000），每个干草垛都在不同的点上。为了让每个干草垛之间都隔着一定的距离，请你回答农夫John的Q个问题（1<=Q<=100,000），每个问题都会给出一个范围，询问在这个范围内有多少个干草垛。

（其实就是有一条数轴上有N个不重复的点，再问Q个问题，每个问题是给出一个范围，问此范围内有多少个点？）

（在给出范围的边界上也算）

输入输出格式

输入格式：

The first line contains NN and QQ .

The next line contains NN distinct integers, each in the range 0 \ldots 1,000,000,0000…1,000,000,000 , indicating that there is a haybale at each of those locations.

Each of the next QQ lines contains two integers AA and BB (0 \leq A \leq B \leq 1,000,000,0000≤A≤B≤1,000,000,000 ) giving a query for the number of haybales between AA and BB , inclusive.

第一行包括N和Q

第二行有N个数字，每个数字的范围在0~1,000,000,000，表示此位置有一个干草垛。

接下来的Q行，每行包括两个数字，A和B（0<=A<=B<=1,000,000,000）表示每个询问的范围

输出格式：

You should write QQ lines of output. For each query, output the number of haybales in its respective interval.

总共Q行，每行输出此范围内的干草垛数量

输入输出样例

输入样例#1：复制

输出样例#1：复制

思路：区间求和难度：普及/提高-

暴力：线段树。只能过一个点，因为测试数据太大（10⁹），而线段树最多能开到10⁷

#include<cstdio>
#define N 10000005
using namespace std;
int n, m, x, y;
struct nond {int ll, rr;int sum;
}tree[4*N];
void up(int now) {tree[now].sum = tree[now*2].sum+tree[now*2+1].sum;
}
void build(int now, int l, int r) {tree[now].ll = l; tree[now].rr = r;if(l == r) {tree[now].sum = 0;return ;}int mid = (l+r) / 2;build(now*2, l, mid);build(now*2+1, mid+1, r);up(now);
}
void change(int now, int s) {if(tree[now].ll == tree[now].rr) {tree[now].sum ++;return ;}int mid = (tree[now].ll+tree[now].rr) / 2;if(s<=mid) change(now*2, s);else change(now*2+1, s);up(now);
}
int query(int now, int l, int r) {if(tree[now].ll==l && tree[now].rr==r) {return tree[now].sum;}int mid = (tree[now].ll+tree[now].rr) / 2;if(l<=mid && mid<r) return query(now*2, l, mid) + query(now*2+1, mid+1, r);else if(r<=mid) return query(now*2, l, r);else return query(now*2+1, l, r);
}
int main() {int a[100005] = {0}, maxn = -1;scanf("%d%d", &n, &m);for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);if(a[i] > maxn) maxn = a[i];}build(1, 1, maxn);for(int i = 1; i <= n; i++) change(1, a[i]);for(int i = 1; i <= m; i++) {scanf("%d%d", &x, &y);if(x > maxn) { printf("0\n"); continue; }printf("%d\n", query(1, x, y));}return 0;
}

线段树代码

正解：运用 c++中的 lower_bound、upper_bound

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 100010
using namespace std;
int n, m, q, tot;
int pos[MAXN], hash[MAXN];
int main() {scanf("%d%d", &n, &q);for(int i = 1; i <= n; i++) scanf("%d", &pos[i]);sort(pos+1, pos+1+n);for(int i = 1; i <= q; i++) {int x, y; scanf("%d%d", &x, &y);cout << upper_bound(pos+1, pos+n+1, y) - lower_bound(pos+1, pos+n+1, x) << ‘\n’;}return 0;
}

PS：lower_bound(val) ：返回容器中第一个值大于等于val 的元素的 iterator 位置

upper_bound(val) ：返回容器中第一个值大于val 的元素的 iterator 位置

转载于:https://www.cnblogs.com/v-vip/p/8746161.html