I can do it!

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1010    Accepted Submission(s): 471

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

Author

HyperHexagon

题解：题意好难理解啊，好像是给你一些元素，这些元素有两个性质，把这两个性质分成两个集合，然后找A几何中最大值加上B集合中最大值的最小化值；可以用贪心来找。。。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define T_T while(T--)
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=100010;
struct Node{
	int a,b;
	friend bool operator < (Node x,Node y){
		return x.a>y.a;
	}
};
Node dt[MAXN];
int main(){
	int T,N,kase=0;
	SI(T);
	T_T{
		SI(N);
		for(int i=0;i<N;i++)SI(dt[i].a),SI(dt[i].b);
		sort(dt,dt+N);
		int A,B=0,ans=INF;
		for(int i=0;i<N;i++){
			ans=min(ans,B+dt[i].a);
			B=max(dt[i].b,B);
		}
		printf("Case %d: %d\n",++kase,ans);
	}
	return 0;
}