编程之美区间重合判断

一，问题：

    1. 给定一个源区间[x,y]和N个无序的目标区间[x1,y1] [x2,y2] ... [xn,yn]，判断源区间[x,y]是不是在目标区间内。

     2. 给定一个窗口区域和系统界面上的N个窗口，判断这个窗口区域是否被已有的窗口覆盖。

编程之美的解法：首先利用快排没，将区间按照从小到大排序，[2,3][1,2][3,9]-》[1,2][2,3][3,9]；然后对排序的区间进行合并；合并后运用二分法查找。

#include <iostream>
#include <algorithm>
using namespace std;

struct Line{
	int low,high;
	Line(int l=0,int h=0):low(l),high(h){}
	bool operator<(Line & other)
	{
		return low<other.low;
	}
};

const int MAX = 100;
Line lines[MAX];

//注意咱们搜索的数为刚好小于key的那个值
int BinarySearchLower(Line arr[],int len,int target)
{
	int low = 0;
	int high = len-1;

	while(low < high){
		int mid = (low + high)/2;
		if(arr[mid].low>=target)high = mid-1;
		else low = mid + 1;
			
	}
	return high;
}


int main()
{
	int n, k, i, j;  
    
	cin>>n;  // n是目标区间的个数,k是待查询的源区间的个数  
	for (i=0; i<n; i++)  
		cin >> lines[i].low >> lines[i].high;  

	sort(lines, lines+n);  
 
	int cnt = 0; //合并以后的区间数
	int lasthigh = lines[0].high;  

        //合并区间
	for(i=1; i<n; i++){
		if(lasthigh>=lines[i].low)
			lasthigh = lines[i].high;
		else{
			lines[cnt].high = lasthigh;
			lines[++cnt].low = lines[i].low;
			lasthigh = lines[i].high;
		}

	}
	lines[cnt++].high = lasthigh;

	Line search = Line(1,6);

	int sLow = BinarySearchLower(lines,cnt,search.low);
	int sHigh = BinarySearchLower(lines,cnt,search.high);
	if(sLow==sHigh && search.high<=lines[sLow].high)//注意要判断
		cout<<"Yes"<<endl;
	else cout<<"No"<<endl;

	system("pause");
	return 0;
}