The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are1 ≤ i < j ≤ n and x', y', such that:
- y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
- y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
- x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
4
1 2
1 2
1 0
0 1
0 2
NO
2
1 3
1 0
-1 3
YES
2
1 3
1 0
0 2
YES
2
1 3
1 0
0 3
NO
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
好吧。。。人蠢。。所以没想出来。。
其实画个图就能明白。。。如果两条直线相交在x1,x2之间。。
那么两条直线，分别于x1,x2的交点。。一定是一对逆序对。。。
我们可以得到每条直线在x1,x2的交点的纵坐标。。
然后按照与x1交点的纵坐标降序排。
另外一个比较重要的的是，判断是否有逆序对，只需要判断相邻的就行了。
可以用如下证明（不知道有什么更容易想的办法？）
假设直线i与直线i+k(k>=2)相交，那么直线i与直线i+j(1=<j<k)不相交。
那么li[i].sec<li[i+j].sec,li[i].sec>li[i+k].sec
所以li[i+j].sec>li[i].sec>li[i+k].sec
那么直线i+j一定与i+k相交
当j=k-1时，i+j与i+k相邻。
也就是说...如果任意直线i与j相交...我们总可以转化成一对相邻的直线相交。
因此只判断相邻直线的相交就可以。
画图也能看出来。
最后一点是，要开long long 。
1 /************************************************************************* 2 > File Name: code/cf/#329/B.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年11月05日 星期四 15时34分05秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #include<cctype> 21 22 #define lson l,m,rt<<1 23 #define rson m+1,r,rt<<1|1 24 #define ms(a,x) memset(a,x,sizeof(a)) 25 using namespace std; 26 const int dx4[4]={1,0,0,-1}; 27 const int dy4[4]={0,-1,1,0}; 28 typedef long long LL; 29 typedef double DB; 30 const int inf = 0x3f3f3f3f; 31 const int N=1E5+7; 32 int n; 33 LL x1,x2; 34 pair<LL,LL>li[N]; 35 36 37 38 int main() 39 { 40 #ifndef ONLINE_JUDGE 41 // freopen("in.txt","r",stdin); 42 #endif 43 44 scanf("%d",&n); 45 scanf("%I64d %I64d",&x1,&x2); 46 LL b,k; 47 for ( int i = 0 ; i < n ; i++) 48 { 49 scanf("%I64d %I64d",&k,&b); 50 li[i]=make_pair(k*x1+b,k*x2+b); 51 } 52 sort(li,li+n); 53 bool ok = false; 54 for ( int i = 1 ; i < n ; i++) 55 if (li[i-1].second>li[i].second) 56 { 57 ok = true; 58 break; 59 } 60 61 if (!ok) puts("NO"); 62 else puts("YES"); 63 64 65 66 #ifndef ONLINE_JUDGE 67 fclose(stdin); 68 #endif 69 return 0; 70 }