2191: 【USACO】Crowded Cows

2024-05-13 22:11:24

2191: 【USACO】Crowded Cows

时间限制: 1.000 Sec 内存限制: 64 MB
提交: 26 解决: 19
[命题人:][下载数据: 90]

提交状态报告

题目描述

Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(i) and has height h(i) (1 <= x(i),h(i) <= 1,000,000,000). A cow feels "crowded" if there is another cow at least twice her height within distance D on her left, and also another cow at least twice her height within distance D on her right (1 <= D <= 1,000,000,000). Since crowded cows produce less milk, Farmer John would like to count the number of such cows. Please help him.

输入

* Line 1: Two integers, N and D. * Lines 2..1+N: Line i+1 contains the integers x(i) and h(i). The locations of all N cows are distinct.

输出

* Line 1: The number of crowded cows.

样例

输入复制

6 4 10 3 6 2 5 3 9 7 3 6 11 2

输出复制

2

提示

INPUT DETAILS: There are 6 cows, with a distance threshold of 4 for feeling crowded. Cow #1 lives at position x=10 and has height h=3, and so on. OUTPUT DETAILS: The cows at positions x=5 and x=6 are both crowded.

来源/分类

USACO 2013 November Silver

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;struct Cow {long long x , hi;bool operator < (const Cow &a) const {return hi > a.hi;}
};
Cow tag[50005];
long long n;
long long d , ans = 0;long long myabs(long long x) {return x>0?x:(-x);
}int main() {scanf("%d%lld",&n,&d);for(long long i=1; i<=n; i++)scanf("%lld%lld",&tag[i].x,&tag[i].hi);sort(tag+1 , tag+n+1);for(long long i=1; i<=n; i++) {long long flag1 = 0 , flag2 = 0;for(long long j=1; j<i; j++) {if(tag[j].hi < tag[i].hi*2)break;if(myabs(tag[j].x-tag[i].x) <= d) {if(tag[j].x > tag[i].x)flag1 = 1;if(tag[j].x < tag[i].x)flag2 = 1;if(flag1 && flag2) {ans++;break;}}}   }printf("%lld",ans);return 0;
}

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