hdu 3552 I can do it! （贪心）

题 目 ：

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

 

参 考：http://blog.csdn.net/dgq8211/article/details/7748078

题 意：

有n个物品，每个物品有两种属性X和Y。问怎样把这些物品分成两个集合(可空)，使得 集合A中的max(X)+集合B中的max(Y)  的值最小。

 代 码 ：

#include<iostream>
#include<algorithm>
#include<stdio.h>

using namespace std;

struct food
{
    int x,y;

}F[100005];

bool com( food a,food b)
{
    return  a.x>b.x ;
}

int main( ) {

    int N,n,Min_sum,Max_y,x=0;
    cin >>N;
    while( N-- ){
        cin >>n;
        for( int i=1;i<=n;i++)
            scanf("%d %d",&F[i].x,&F[i].y);
       // if( n==1 ) cout <<"Case " <<++x <<": " <<min(F[1].x,F[1].y) <<endl;
       //  这句话 要不要都正确 说明 数据中没有这组特殊数据 ；
       // else {
            sort(F+1,F+n+1,com);
            Min_sum=F[1].x; Max_y=F[1].y;

            for( int i=1;i<=n;i++) {

                Min_sum= min( Min_sum, F[i].x+Max_y);  //  感觉像排完序之后 ， 用隔板隔开 ，看哪种情况是最优的 。
                Max_y= max( Max_y, F[i].y);
            }
            cout <<"Case " <<++x <<": " <<Min_sum <<endl;
       // }
    }
    return 0;
}