1074 Reversing Linked List （PAT甲级）

#include 
#include 
#include 
const int MAXN = 100001;

struct node{
    int data;
    int next;
};

int curr, N, K, address;
node vec[MAXN];
std::vector ans;

int main(){
    scanf("%d %d %d", &curr, &N, &K);
    for(int i = 0; i < N; ++i){
        scanf("%d", &address);
        scanf("%d %d", &vec[address].data, &vec[address].next);
    }
    while(curr != -1){
        ans.push_back(curr);
        curr = vec[curr].next;
    }
    for(int i = 0; i < ans.size() / K; ++i){
        reverse(ans.begin() + i * K, ans.begin() + (i + 1) * K);
    }
    for(int i = 0; i < ans.size() - 1; ++i){
        printf("%05d %d %05d\n", ans[i], vec[ans[i]].data, ans[i + 1]);
    }
    printf("%05d %d -1\n", ans.back(), vec[ans.back()].data);
    return 0;
}

题目如下：

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