PAT (Advanced Level) Practise 1109. Group Photo (25) 结构体排序
来源:互联网 发布:大学生犯罪知乎 编辑:程序博客网 时间:2024/04/29 03:26
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 3Tom 188Mike 170Eva 168Tim 160Joe 190Ann 168Bob 175Nick 186Amy 160John 159Sample Output:
Bob Tom Joe NickAnn Mike EvaTim Amy John
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define PAUSE system("pause")using namespace std;const int maxn = 10000 + 10;struct node {int h;char name[10];bool operator < (const node& on) const {if (on.h == h) {//return (int)name[0] < (int)on.name[0]; //debug 总是犯这种SB的错误(-.-)!return strcmp(name, on.name) < 0;}else {return h > on.h;}}}p[maxn], ans[maxn];int N, K;void out(int len) {for (int i = 1; i <= len; i++) {printf(i == 1 ? "%s" : " %s", ans[i].name);}puts("");}int main(){int row, lr, len, mid, cnt, i, j, cou;scanf("%d%d", &N, &K);row = N / K;lr = row + (N - row * K);if (N < K) {row = lr = 1;}for (int i = 1; i <= N; i++) {scanf("%s%d", p[i].name, &p[i].h);}sort(p + 1, p + 1 + N);cou = 1;for (int k = 1; k <= K; k++) {j = -1; i = 1;mid = (k == 1 ? lr : row) / 2 + 1;ans[mid] = p[cou++];cnt = 1;len = (k == 1 ? lr : row);while (cnt <= len && (mid + j >= 1 || mid + i <= len)) {if (mid + j >= 1) {ans[mid + j] = p[cou++];cnt++;j--;}if (mid + i <= len) {ans[mid + i] = p[cou++];cnt++;i++;}}out(cnt);}return 0;}
- PAT (Advanced Level) Practise 1109. Group Photo (25) 结构体排序
- PAT (Advanced Level) Practise 1109 Group Photo (25)
- PAT (Advanced Level) Practise 1109 Group Photo (25)
- 【PAT】【Advanced Level】1109. Group Photo (25)
- 1109. Group Photo (25)[结构体排序]
- PAT (Advanced Level) 1025. PAT Ranking (25) 结构体排序
- 1029. Median (25)【排序】——PAT (Advanced Level) Practise
- [PAT] (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) Practise
- PAT (Advanced Level) 1028. List Sorting (25) 结构体排序
- 快速排序算法实现
- 让键盘消失的方法
- [Powershell]导出特定AD用户的属性
- 4.23 GDOI赛前模拟 总结
- 编程之美之第k大数
- PAT (Advanced Level) Practise 1109. Group Photo (25) 结构体排序
- java--toString
- 单调队列Monotonic Queue
- 找出对oracle测试库的表执行delete的用户
- hdu 5247 找连续数-2015年百度之星程序设计大赛 - 初赛(1)
- UcosII 学习
- iPhone设备型号列表
- LeetCode *** 139. Word Break
- hadoop学习序曲之java基础篇--java内部类 异常