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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:
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
题意：给你x,y,还有n条直线，问你在x到y的区域内部是否存在交点，不包括x,y上
题解：贪心，我们知道在x,y上必然存在两个交点，所以我们贪心按照左边从小到大排序，如果右边不是递增的则存在交点
///1085422276 #include<bits/stdc++.h> using namespace std ; typedef long long ll; #define mem(a) memset(a,0,sizeof(a)) inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) { if(ch==‘-‘)f=-1; ch=getchar(); } while(ch>=‘0‘&&ch<=‘9‘) { x=x*10+ch-‘0‘; ch=getchar(); } return x*f; } //**************************************** #define maxn 1000000+5 #define mod 1000000007 double k[maxn],b[maxn]; struct ss{ double l; double r; int index; }G[maxn]; int cmp(ss s1,ss s2){ if(s1.l==s2.l)return s1.r<s2.r; else return s1.l<s2.l; } int n; int main(){ n=read(); double x,y; scanf("%lf%lf",&x,&y); int kk=0; for(int i=1;i<=n;i++){ scanf("%lf%lf",&k[i],&b[i]); double xx=k[i]*x+b[i]; double yy=k[i]*y+b[i]; G[++kk].l=xx;G[kk].r=yy; } sort(G+1,G+kk+1,cmp); double now=-1,last=G[1].r; for(int i=2;i<=kk;i++){ if(G[i].r<last){ cout<<"YES"<<endl; return 0; } now=G[i].l; last=G[i].r; } cout<<"NO"<<endl; return 0; }
Codeforces Round #329 (Div. 2)B. Anton and Lines 贪心
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原文地址：http://www.cnblogs.com/zxhl/p/4939104.html
