【codeforces 766B】Mahmoud and a Triangle

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Mahmoud has n line segments, the i-th of them has length ai. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn’t accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.

Mahmoud should use exactly 3 line segments, he can’t concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.

Input
The first line contains single integer n (3 ≤ n ≤ 105) — the number of line segments Mahmoud has.

The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 109) — the lengths of line segments Mahmoud has.

Output
In the only line print “YES” if he can choose exactly three line segments and form a non-degenerate triangle with them, and “NO” otherwise.

Examples
input
5
1 5 3 2 4
output
YES
input
3
4 1 2
output
NO
Note
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.

【题目链接】:http://codeforces.com/contest/766/problem/B

【题意】

给你n条边;
问你能不能从中挑出3条边来组成一个三角形;(只要是三角形就行)

【题解】

可以这样;
设
c<=b<=a
先把边的长度升序排;
如果是三角形那么；
a+b>c
a+c>b即c>b-a
b+c>a即c>a-b
这里b-a是个负数,所以不用理他
只考虑a-b就好;
而想让a-b最小,肯定是选长度相邻的两条边;
这样从大到小,维护i..n这个区间范围内a-b的最小值mi;
看看a[i]是不是大于mi;一旦有符合的就ok了;
(至于a+b>c,因为a和b都在i..n这个范围内选,所以a+b>c肯定成立的)

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1e5+100;

int a[MAXN],n;

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]);
    sort(a+1,a+1+n);
    int d = a[n]-a[n-1];
    rep2(i,n-2,1)
        {
            if (d<a[i])
            {
                puts("YES");
                return 0;
            }
            d = min(d,a[i+1]-a[i]);
        }
    puts("NO");
    return 0;
}