【lightoj1307】Counting Triangles

最新推荐文章于 2020-05-17 00:02:50 发布

原创最新推荐文章于 2020-05-17 00:02:50 发布 · 424 阅读

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#二分优化

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本文介绍了一种通过二分查找优化的算法，用于解决给定不同长度棍子后能组成多少个有效三角形的问题。文章提供了一个具体的C++实现案例，通过排序和二分查找技术来高效求解。

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J - J 圆周率用acos(-1.0) 使用longlong

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1307

uDebug

Description

You are given N sticks having distinct lengths; you have to form some triangles using the sticks. A triangle is valid if its area is positive. Your task is to find the number of ways you can form a valid triangle using the sticks.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer N (3 ≤ N ≤ 2000). The next line contains N integers denoting the lengths of the sticks. You can assume that the lengths are distinct and each length lies in the range [1, 10⁹].

Output

For each case, print the case number and the total number of ways a valid triangle can be formed.

Sample Input

3 12 5 4 9

1 2 3 4 5 6

100 211 212 121

Sample Output

Case 1: 3

Case 2: 7

Case 3: 4

比赛时没TLE豁然开朗，原来用二分可以轻松解决超时问题，看来二分更多的时候是用来优化程序的，我喜欢这种题。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int a[2005];
bool judge(int mid,int i,int j) {
	if(a[i]+a[j]>a[mid]&&a[mid]-a[i]<a[j]) {
		return true;
	}
	return false;
}
int main() {
	int T,p=0;
	scanf("%d",&T);
	while(T--) {
		int n;
		scanf("%d",&n);
		for(int i=0; i<n; i++) {
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		int ans=0;
		for(int i=0; i<n; i++) {
			for(int j=i+1; j<n; j++) {
				int cnt=0;
				int l=j+1,r=n-1;
				while(l<=r) {
					int mid=l+r>>1;
					if(judge(mid,i,j)) {
						cnt=mid;
						l=mid+1;
					} else {
						r=mid-1;
					}
				}
				if(cnt)
					ans=ans+cnt-j;
			}
		}
		printf("Case %d: %d\n",++p,ans);
	}
	return 0;
}