hdoj I can do it! 3552 （贪心&模拟）

I can do it!

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1016 Accepted Submission(s): 473

Problem Description

Given n elements, which have two properties, say Property A and Property B. For convenience, we use two integers A _i and B _i to measure the two properties.
Your task is, to partition the element into two sets, say Set A and Set B , which minimizes the value of max _{(x∈Set A)} {A _x}+max _{(y∈Set B)} {B _y}.
See sample test cases for further details.

Input

There are multiple test cases, the first line of input contains an integer denoting the number of test cases.
For each test case, the first line contains an integer N, indicates the number of elements. (1 <= N <= 100000)
For the next N lines, every line contains two integers A _i and B _i indicate the Property A and Property B of the ith element. (0 <= A _i, B _i <= 1000000000)

Output

For each test cases, output the minimum value.

Sample Input

1
3
1 100
2 100
3 1

Sample Output

Case 1: 3

//题意： /*有n个元素，每个元素有两个属性（x,y）。现在要将这n个元素分成两个集合（A,B）; 分成两个集合后再求出最小的sum的值， 其中sum=（集合A中最大的属性值x）+（集合B中最大的属性值y）; */

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct zz
{
	int x;
	int y;
}q[100010];
int cmp(zz a,zz b)
{
	return a.x>b.x;
}
int main()
{
	int t,T=1,n;
	int i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%d%d",&q[i].x,&q[i].y);
		sort(q+1,q+n+1,cmp);
		int ans=0x3f3f3f3f;
		q[n+1].x=0;
		q[0].y=0;
		int B=0;
		for(i=1;i<=n+1;i++)
		{
			B=max(B,q[i-1].y);
			ans=min(ans,q[i].x+B);
		}
		printf("Case %d: %d\n",T++,ans);
	}
	return 0;
}