PAT 1109 Group Photo[仿真][难]

Posted 2020-12-31 bluebluesea

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1109 Group Photo（25 分）

Formation is very important when taking a group photo. Given the rules of forming

The number of people in each row must be
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
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

Output Specification:

For each case, print the formation -- that is, print the names of people in

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：模拟拍合照，假设有N个人需要排K行，有几点要求：

1.每行必须有N/K个人（四舍五入），剩下的人排到最后一行

2.在后一排的所有人不能比首排的人矮

3.每一排中，最高的人站C位（m/2+1），

4.每一排中的站法，先往最高的人右手边站，再往左手边站

5.当身高一样时，按名字字母升序排列。

//猛一看就觉得很难，很多要求，这是个仿真类的题目吧。

//每一行内的人用什么数据结构来存呢?可以左右插入！

#include <iostream>
#include <cstring>
#include<stdio.h>
using namespace std;
struct PE{
    int height;
    char name[10];
}pe[10001],pe2[10001];
int n,k;
bool cmp(PE& a,PE &b){
    if(a.height>b.height)
        return true;
    else if(a.height==b.height&&strcmp(a.name,b.name)>0)
        return true;
    return false;
}
int main() {
    scanf("%d %d",&n,&k);
    char na[10];
    int h;
    for(int i=0;i<n;i++){
        scanf("%s %d",pe[i].name,&pe[i].height);//这里无效赋值。。
    }
    sort(pe,pe+n,cmp);
    int m=n/k;//一共k排，出最后行外每排m个人
    for(int i=0;i<k;i++){
        int st=i*k;
        for(int j=st;j<st+m;j++){
            //这right和left乱放太难控制了吧。
            //感觉实现就在眼前，就是写不出来。。气死了。            
        }
    }

    return 0;
}

View Code

//这是我写的，写到一半写不下去了，因为不会安排。

代码来自：https://www.liuchuo.net/archives/1926

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    string name;
    int height;
};
int cmp(node a, node b) {
    return a.height != b.height ? a.height > b.height : a.name < b.name;
    //这个一句代码，确实比我那个有好几return的好！
}
int main() {
    int n, k, m, i, j;
    cin >> n >> k;
    vector<node> stu(n);
    for(i = 0; i < n; i++)
        cin >> stu[i].name >> stu[i].height;
    sort(stu.begin(), stu.end(), cmp);
    int t = 0, row = k;
    while(row) {
        if(row == k)
            m = n - n / k * (k - 1);//这是在安排最后一行的人数。
        else
            m = n / k;
        vector<string> stemp(m);//每一行都重新定义一个向量。
        stemp[m / 2] = stu[t].name;//这个中间值很厉害了，
        //题目中给的是m/2+1，但是题目中是从1开始计数，数组中下标从0开始。
        // 左边一列
        j = m / 2 - 1;
        for(i = t + 1; i < t + m; i = i + 2)
            stemp[j--] = stu[i].name;
        // 右边一列
        j = m / 2 + 1;
        for(i = t + 2; i < t + m; i = i + 2)
            stemp[j++] = stu[i].name;
        // 输出当前排
        cout << stemp[0];
        for(i = 1; i < m; i++) cout << " " << stemp[i];
        cout << endl;
        t = t + m;
        row--;
    }
    return 0;
}