hdoj--3552--I can do it!(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): 1022    Accepted Submission(s): 475

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

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

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#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	int a,b;
}num[1000100];
int cmp(node s1,node s2)
{
	return s1.a>s2.a;
}
int main()
{
	int t,n;
	scanf("%d",&t);
	int k=1;
	while(t--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		scanf("%d%d",&num[i].a,&num[i].b);
		int maxx1=0,ans=0x3f3f3f3f;
		sort(num+1,num+1+n,cmp);
		num[0].b=0;
		num[n+1].a=0;
		for(int i=1;i<=n+1;i++)
		{
			maxx1=max(maxx1,num[i-1].b);
			ans=min(ans,num[i].a+maxx1);
		}
		printf("Case %d: %d
",k++,ans);
	}
	return 0;
}