HDU 3552 I can do it! 【贪心】

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

/*
题目大意：一个元素分为2个属性，a和b。然后将元素分成2个集合A和B。求出A集合的a属性的最大值+B集合的b属性的最大值，的最小值 
解题思路：一篇很详细的结题报告
http://blog.csdn.net/dgq8211/article/details/7748078 
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct Set
{
    int a, b;
}SS[100002];
int cmp(Set x, Set y)
{
    return x.a>y.a;
}
int main()
{
    int T, n, k=0, i;
    scanf("%d", &T);
    while(T--)
    {
        k++;
        scanf("%d", &n);
        for(i=0; i<n; i++)
        {
            scanf("%d%d", &SS[i].a, &SS[i].b);
        }
        sort(SS, SS+n, cmp);
        int maxsum=SS[0].a, maxb=SS[0].b;
        for(i=1; i<n; i++)
        {
            maxsum=min(maxsum, SS[i].a+maxb);
            maxb=max(maxb, SS[i].b);
        }
        printf("Case %d: %d\n", k, maxsum);
    }
    return 0;
}