pat 1074 Reversing Linked List (25)

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.

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 (<= 105) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist
to be reversed. 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 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
using namespace std;

struct point
{
char add[20], cnext[20];
int data;
int idx;
} num[100010];

bool cmp(const point &a1, const point &a2) {
return a1.idx < a2.idx;
}

int main() {
char add[20];
int n, k;
map<string, int> mapx;

scanf("%s %d %d", add, &n, &k);

for (int i = 0; i < n; ++i)
{
scanf("%s %d %s", num[i].add, &num[i].data, num[i].cnext);
num[i].idx = 0x7fffffff;
mapx.insert(make_pair(num[i].add, i));
}

int x = mapx.find(add)->second, i = 0;

for(i = 0; i < n; ) {
num[x].idx = i;
i++;

if(strcmp(num[x].cnext, "-1") == 0) break;

x = mapx.find(num[x].cnext)->second;
}

sort(num, num+n, cmp);

n = i;
int high = n/k*k;

for(i = 0; i+k <= high; i=i+k) {

for(int j = k-1; j > 0; j--) {
printf("%s %d %s\n", num[i+j].add, num[i+j].data, num[i+j-1].add);
}

printf("%s %d ", num[i].add, num[i].data);

if(i+k==n) {
printf("-1\n");
} else if(i+2*k < n) {
printf("%s\n", num[i+2*k-1].add);
} else if(i+2*k > n) {
printf("%s\n", num[i+k].add);
}
}

for(; i < n; i++) {
printf("%s %d %s\n", num[i].add, num[i].data, num[i].cnext);
}
return 0;
}