A1109. Group Photo

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 (<=10000), 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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
typedef struct{
    string name;
    int high;
}info;
info people[10001], line[10001];
bool cmp(info a, info b){
    if(a.high != b.high)
        return a.high > b.high;
    else return a.name < b.name;
}
int main(){
    int N, K;
    cin >> N >> K;
    for(int i = 0; i < N; i++){
        cin >> people[i].name >> people[i].high;
    }
    sort(people, people + N, cmp);
    int lineSize = N / K;
    int lastSize = N - lineSize * (K - 1);
    int index = 0;
    for(int i = 0; i < K; i++){
        int len = i == 0 ? lastSize : lineSize;
        int mid = len / 2;
        int left = mid - 1, right = mid + 1;
        line[mid] = people[index++];
        while(left >= 0 && right < len){
            line[left--] = people[index++];
            line[right++] = people[index++];
        }
        if(left == 0) 
            line[left--] = people[index++];
        for(int j = 0; j < len - 1; j++){
            cout << line[j].name << " ";
        }
        cout << line[len - 1].name << "
";
    }
    cin >> N;
    return 0;
}

总结：

1、题意：按照集体照像的原则排队，每一排站N / K个人，最后一排如果有多的则都站最后一排。每一排都比前一排要高。对于每一排，个子最高的站中间，然后次高的站他的左边，第三高的站他的右边。如此从中间向两边一左一右的安排。 如果有一样高的，则按照姓名字典序排序。

2、本题可以先整体从高到低排序。然后逐行安排，每排完一行就输出该行。使用index作为整体数组的指针，每排完一个人就+1。对于每一排，从中间向两边安排即可。

3、最好不要从前往后排，这样需要保存后统一输出，很麻烦。