1028 List Sorting (25分)
Excel can sort records according to any column. Now you are supposed to imitate this function.
Input Specification:
Each input file contains one test case. For each case, the first line contains two integers N (≤10
?5
??) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).
Output Specification:
For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.
Sample Input 1:
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
Sample Output 1:
000001 Zoe 60
000007 James 85
000010 Amy 90
Sample Input 2:
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
Sample Output 2:
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
Sample Input 3:
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
Sample Output 3:
000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90
小结
没啥难度，基本排序，一次过
#include <iostream>
#include <vector>
using namespace std;typedef struct
{int length=999;vector<int> children;
}Node;Node node[101];
void writeLen(int nodeNum, int Len) {if (node[nodeNum].length < Len) return;else node[nodeNum].length = Len;for (int i = 0; i < node[nodeNum].children.size(); i++){writeLen(node[nodeNum].children[i], Len + 1);}}int main() {int N, M;//N树的节点 M非叶子的节点数量cin >> N >> M;if (N == 0) {cout << 0;return 0;}for (int i = 0; i < M; i++){int root, number;cin >> root >> number;for (int j = 0; j < number; j++){int child;cin >> child;node[root].children.push_back(child);}}writeLen(1, 0);int maxLen=0;for (int i = 0; i < 101; i++){if (node[i].length > maxLen&& node[i].length!=999) {maxLen = node[i].length;}}for (int i = 0; i < maxLen+1; i++){int num = 0;if (i != 0)cout << " ";for (int j = 0; j < 101; j++){if (node[j].length == i && node[j].children.size() == 0)num += 1;}cout << num ;}return 0;
}