solution Of Pat 1109. Group Photo (25)

1109.Group Photo (25)

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

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结题思路：

题意要求我们对每个人的身高进行排序，然后对于局部数据规格化输出。
要求1：CountPerRow=N/K，前K-1行人数CountPerRow，最后一行人数CountPerRow+N%K；
要求2：后一排最矮的也比前一排最高的要高；
要求3：每排最高的站在中间位置，每排从0开始计数的话，即最高的人的位置在CountPerRow/2处；
要求4：最下来的人交替在左边和右边排序。（可通过距离最高的那个人的距离值得奇偶性直接进行判断）；
要求5：当高度相同，我们按字母序递增排序，验证数据保证当前情况下的名字绝对不会重复。

程序步骤：
第一步我们要按高度和字母序对输入数据进行排序；
第二步对最后一行数据单独处理；
第三步处理前面的数据（重复操作）。

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具体程序(AC)如下：

#include <iostream>#include <algorithm>#include <vector>#include <cmath>using namespace std;struct person{    string name;    int height;};int cmp(const person& a,const person& b){    if(a.height!=b.height)        return a.height<b.height;    else        return a.name>b.name;}void reconstruct(vector<person>& A,vector<string>& B,int start,int len){    if(len==0)return;    int mid=len/2;    for(int i=0;i<len;++i)    {        int dif=abs(len-1-i);        if(dif%2)//距离为奇数的部分，left        {            B[mid-(dif+1)/2]=A[i+start].name;        }        else//距离为偶数的部分，right        {            B[mid+dif/2]=A[i+start].name;        }    }    cout<<B[0];    for(int i=1;i<len;++i)    {        cout<<" "<<B[i];    }}int main() {    // your code goes here    int n,rows;    vector<person> pL;    pL.clear();    cin>>n>>rows;    person a;    for(int i=0;i<n;++i)    {        cin>>a.name>>a.height;        pL.push_back(a);    }//输入结束    sort(pL.begin(),pL.end(),cmp);    int CountPerRow,lastRow;    CountPerRow=n/rows;    lastRow=n%rows+CountPerRow;    vector<string> nameList;    nameList.resize(lastRow);    reconstruct(pL,nameList,n-lastRow,lastRow);    for(int i=rows-2;i>=0;--i)    {        cout<<endl;        reconstruct(pL,nameList,i*CountPerRow,CountPerRow);    }    return 0;}