洛谷 P3184 [USACO16DEC]Counting Haybales数草垛

   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 100,0001N100,000 ) 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 100,0001Q100,000 ), 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 1,000,000,00001,000,000,000 , 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 1,000,000,0000AB1,000,000,000 ) 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
4 6
3 2 7 5
2 3
2 4
2 5
2 7
4 6
8 10
输出样例#1: 复制
2
2
3
4
1
0

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;  
}
AC

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

posted @ 2018-04-08 17:17  落云小师妹  阅读(298)  评论(0编辑  收藏  举报