【PAT】B1055 集体照(25 分)

很简单的two points问题

注意:K是行数

#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
struct people{
	char name[10];
	int height;
}arr[10005];
bool cmp(people A,people B){
	if(A.height!=B.height) return A.height<B.height;
	else return strcmp(A.name,B.name)>0;
}
int main(){
	int N,K;scanf("%d %d",&N,&K);
	for(int i=0;i<N;i++){//输入
		getchar();//吸收上一行的换行符
		scanf("%s %d",arr[i].name,&arr[i].height);
	}
	sort(arr,arr+N,cmp);//排序
	//让储存从下标1开始
	people data[N/K*3];
	int flag=N-1;
	for(int i=K;i>=1;i--){//K排
		int left=1,right=N/K;
		if(i==K) right=N/K+N%K;
		int mid=right/2+1;
		int l=mid-1,r=mid+1;
		data[mid]=arr[flag--];
		while(l>=left||r<=right){
			if(l>=left){
				data[l--]=arr[flag--];
			}
			if(r<=right){
				data[r++]=arr[flag--];
			}

		}
		for(int j=1;j<=right;j++){
			if(j!=1) printf(" ");
			printf("%s",data[j].name);
			if(j==right&&i!=1) printf("
");
		}
	}
	getchar();
	return 0;
}

原文地址:https://www.cnblogs.com/hebust/p/9496797.html