PAT_甲级_1109 Group Photo (25point(s)) (C++)【模拟】

目录

​​1，题目描述​​

​​ 题目大意​​

​​2，思路​​

​​数据结构​​

​​算法​​

​​3，AC代码​​

​​4，解题过程​​

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1，题目描述

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

将N个人按照规则排成K行站队。规则如下：

“我”面向队伍，队伍第一排最矮，往后逐渐增高；

• 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：每一行的人数为N/K向下取整（正常除即可），多出来的人全部站在最后一行；
• All the people in the rear row must be no shorter than anyone standing in the front rows：后一排的比前一排的任何一个人高；
• 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)：最高的人站的位置，假如一行有3个人，编号0-2，tallest站在3/2=1，假如一行有4个人，tallest站在4/2=2；
• 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 ：“我”面向队伍，4th 2th 1th 3th    /    2th 1th 3th；
• When there are many people having the same height, they must be ordered in alphabetical (increasing) order of their names：当身高相等时，名字更“大”的同学更“矮”；

2，思路

数据结构

• struct node{
  string name;
  int height;
  }data[10006];所有人的身高和名字数据；
• node infor[K][lenRear]：队伍站位；

算法

2. 接收数据至data：



4. 从低到高排序：



6. 记录每一行最低的人的位置（便于定位最高的人tallest），每次排一行（一行中左右交替赋值），从前往后：



8. 输出

 

3，AC代码

#include<bits/stdc++.h>
using namespace std;
struct node{
    string name;
    int height;
}data[10006];
bool cmp1(node a, node b){
    if(a.height != b.height)
        return a.height < b.height;
    else
        return a.name > b.name;                     //注意排序规则
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    int N, K;
    scanf("%d %d", &N, &K);
    node n;
    for(int i = 0; i < N; i++){
        cin>>n.name>>n.height;
        data[i] = n;
    }
    sort(data, data + N, cmp1);
    int len = N / K, lenRear = len + N % K, index = 0;//index每一行的最低的人
    node infor[K][lenRear];
    int tallest, left, right;           //left左指针 right右指针 分别指向将要赋值的位置
    bool flag;                          //判断左右 false左 true右
    for(int i = 0; i < K; i++){         //K行
        if(i != K - 1){
            tallest = index + len - 1;  //当前行最高的人在data中的编号
            left = len / 2 - 1;right = len / 2;
        }else{
            tallest = N - 1;
            left = lenRear / 2 - 1;right = lenRear / 2;
        }
        flag = true;                    //右1th 左2th 右3th......右左交替!!!
        for(int j = 0; j < (i==K-1 ? lenRear:len); j++){
            if(flag){
                infor[i][right++] = data[tallest--];
                flag = false;
            }else{
                infor[i][left--] = data[tallest--];
                flag = true;
            }
        }
        index += len;                   //index每一行的最低的人
    }
    for(int i = K - 1; i >= 0; i--){
        for(int j = 0; j < (i==K-1 ? lenRear:len); j++){
            if(j == 0)
                cout<<infor[i][j].name;
            else
                cout<<' '<<infor[i][j].name;
        }
        cout<<endl;
    }
    return 0;
}

4，解题过程

一发入魂