PTA 1165 Block Reversing (25 分)
1165 Block Reversing (25 分)
Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10
5
) which is the total number of nodes, and a positive K (≤N) which is the size of a block. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218
Sample Output:
71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1
感悟
不难的一道结构体排序题,但是又卡了一个点,经典的PTA单链表有无用结点.结构体用大括号整体赋值的时候,按顺序赋值,未赋值元素值为0,这个BUG卡了有20分钟.
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
typedef pair<int, int> PII;
struct Node {
int address;
int data;
int next;
PII x;
bool operator < (const Node& w) const {
return x < w.x;
}
}a[N];
int main()
{
int start, n, k;
scanf("%d%d%d", &start, &n, &k);
for(int i = 0 ; i < N ; i ++) a[i].x = {N,N};
for (int i = 0; i < n; i++)
{
int address,data,next;
scanf("%d%d%d", &address, &data, &next);
a[address] = {address, data, next, {N, N}};
}
int cnt = 0;
for (int i = start,j = 0; i != -1; i = a[i].next, j ++, j %= k, cnt ++)
a[i].x = {n/k - cnt/k,j};
sort(a,a + N);
// 会有不在链表里的结点
for(int i = 0 ; i < cnt ; i ++)
{
if(i != cnt - 1) printf("%05d %d %05d\n",a[i].address,a[i].data,a[i+1].address);
else printf("%05d %d -1\n",a[i].address,a[i].data);
}
return 0;
}