1109 Group Photo (25 point(s))-优快云博客

本文介绍了一种算法，用于解决群体照拍摄时的人员排列问题，确保后排人员不低于前排，且根据身高和姓名顺序合理分布。

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Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following:

The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in the last row;
All the people in the rear row must be no shorter than anyone standing in the front rows;
In each row, the tallest one stands at the central position (which is defined to be the position (m/2+1), where m is the total number of people in that row, and the division result must be rounded down to the nearest integer);
In each row, other people must enter the row in non-increasing order of their heights, alternately taking their positions first to the right and then to the left of the tallest one (For example, given five people with their heights 190, 188, 186, 175, and 170, the final formation would be 175, 188, 190, 186, and 170. Here we assume that you are facing the group so your left-hand side is the right-hand side of the one at the central position.);
When there are many people having the same height, they must be ordered in alphabetical (increasing) order of their names, and it is guaranteed that there is no duplication of names.

Now given the information of a group of people, you are supposed to write a program to output their formation.

Input Specification:

Each input file contains one test case. For each test case, the first line contains two positive integers N (≤104), the total number of people, and K (≤10), the total number of rows. Then N lines follow, each gives the name of a person (no more than 8 English letters without space) and his/her height (an integer in [30, 300]).

Output Specification:

For each case, print the formation -- that is, print the names of people in K lines. The names must be separated by exactly one space, but there must be no extra space at the end of each line. Note: since you are facing the group, people in the rear rows must be printed above the people in the front rows.

Sample Input:

10 3
Tom 188
Mike 170
Eva 168
Tim 160
Joe 190
Ann 168
Bob 175
Nick 186
Amy 160
John 159

Sample Output:

Bob Tom Joe Nick
Ann Mike Eva
Tim Amy John

题目大意：给一组人（n个）的姓名和身高，进行排序；

排序要求：

1）每行中的人数必须是N / K（向下舍入到最接近的整数），所有额外的人（如果有的话）都在最后一行;
后排的所有人都不得比站在前排的任何人都短;
2）在每一行中，最高的一个站在中心位置（定义为位置（m / 2 + 1），其中m是该行中的总人数，并且除法结果必须向下舍入到最接近的整数）;
3）在每一行中，其他人必须以不增加的高度顺序进入该行，交替地将他们的位置首先向右，然后向最高位置的左侧（例如，给予五个人的高度为190,188， 186,175和170，最终阵型将分别为175,188,190,186和170.这里我们假设你正面对着这个小组，所以你的左手边是中央位置的左手边位置。）;
4）当有许多人具有相同的高度时，必须按其名称的字母顺序（递增）顺序排序，并且保证不存在重复的名称。
现在给出一组人的信息，你应该写一个程序来输出他们的形成

分析：

对输入数据排序后模拟一下填充过程即可：对于每一行，首先安置最高的，再填充最高的左列，然后填充最高的右列。用vector<string>temp暂存每一组名字，用p游标扫描输入数据进行填充，遍历输出。

完整代码：

#include<bits/stdc++.h>
using namespace std;

struct node{
	string name;
	int height;
	bool operator <(const node & a)const{ // < 重载 
		return height!=a.height? height>a.height:name<a.name;
	} 
};

int main(){
	int n,k,row,cnt,p=0,j=0; // p用于扫描v【】数组 
	cin>>n>>k;
	vector<node>v(n);
	for(int i=0;i<n;i++) cin>>v[i].name>>v[i].height;
	sort(v.begin(),v.end());
	row=k;
        while(row){
    	    row==k? cnt=n-n/k*(k-1):cnt=n/k; //cnt记录每一行人数 
            vector<string>temp(cnt); //临时存储每一行人名 
            temp[cnt/2]=v[p].name; // 
	    j=cnt/2-1;    	
    	    //左列填充 
	    for(int i=p+1;i<p+cnt;i+=2){
		temp[j--]=v[i].name;
	    } //右列填充 
	    j=cnt/2+1;
	    for(int i=p+2;i<p+cnt;i+=2){
		temp[j++]=v[i].name;
    	    } 
	    for(int i=0;i<cnt;i++)
    	        printf("%s%s",temp[i].c_str(),i!=cnt-1? " ":"\n");
	    p+=cnt,row--;
	}
	return 0;
}

That’s all ！