Background There are a lot of monkeys in a mountain. Every one wants to be the monkey king. They keep arguing with each other about that for many years. It is your task to help them solve this problem.
Problem Monkeys live in different places of the mountain. Let a point (x, y) in the X-Y plane denote the location where a monkey lives. There are no two monkeys living at the same point. If a monkey lives at the point (x0, y0), he can be the king only if there is no monkey living at such point (x, y) that x>=x0 and y>=y0. For example, there are three monkeys in the mountain: (2, 1), (1, 2), (3, 3). Only the monkey that lives at the point (3,3) can be the king. In most cases, there are a lot of possible kings. Your task is to find out all of them.
Input
The input consists of several test cases. In the first line of each test case, there are one positive integers N (1<=N<=50000), indicating the number of monkeys in the mountain. Then there are N pairs of integers in the following N lines indicating the locations of N monkeys, one pair per line. Two integers are separated by one blank. In a point (x, y), the values of x and y both lie in the range of signed 32-bit integer. The test case starting with one zero is the final test case and has no output.
Output
For each test case, print your answer, the total number of the monkeys that can be possible the king, in one line without any redundant spaces.
Sample Input
3 2 1 1 2 3 3 3 0 1 1 0 0 0 4 0 0 1 0 0 1 1 1 0
Sample Output
1 2 1
解题思路根据题意知，我们只需要筛选出在右上角的坐标个数即可。 根据x排序，x最大的那个坐标肯定是一个，然后令cur=该坐标的y值。从后往前遍历，查找大于cur的y并且每次更新cur和符合要求的坐标个数。
AC代码：
#include <iostream> #include <algorithm> #include <cstdio> using namespace std; const int MAXN = 50010; struct Node { int x,y; }arr[MAXN]; bool compare(Node &a,Node &b) { if(a.x!=b.x) { return a.x<b.x; } return a.y<b.y; } int main() { int n; while(scanf("%d",&n) && n!=0) { for(int i=0;i<n;i++) { scanf("%d%d",&arr[i].x,&arr[i].y); } sort(arr,arr+n,compare); int num = 1; int cur = arr[n-1].y; for(int i=n-2;i>=0;i--) { if(arr[i].y>cur) { cur = arr[i].y; num++; } } printf("%d\n",num); } return 0; }