I can do it!（贪心）-优快云博客

本文探讨了一种使用贪心算法解决特定问题的方法。给定一组具有两个属性的元素，任务是将这些元素分为两组，使得一组元素属性的最大值与另一组元素属性的最大值之和达到最小。通过分析示例输入输出，我们了解了如何实现这一目标。

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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

Sample Output

Case 1: 3

Author

HyperHexagon

题解：题意好难理解啊，意思是给你一些元素，然后把这些元素分到两个集合里面；

然后找A集合中a最大值加上B集合中b最大值的和的最小化；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;
}

转载于:https://www.cnblogs.com/handsomecui/p/5060965.html