A1074 Reversing Linked List (25 分)

A1074 Reversing Linked List (25 分)

代码长度限制 16 KB

时间限制 400 ms

内存限制 64 MB

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 <vector>
#include <map>

using namespace std;
const int maxn = 100010;
vector<pair<int, int> > temp;

struct Node{
    int address, data, next;
}link[maxn];

int main() {
    int start, N, K;
    int address, data, next;
    scanf("%d%d%d", &start, &N, &K);
    for (int i = 0; i < N; ++i) {
        scanf("%d%d%d", &address, &data, &next);
        link[address].address = address;
        link[address].data = data;
        link[address].next = next;
    }

    address = start;
    while (address != -1){
        temp.push_back(make_pair(link[address].address, link[address].data));
        address = link[address].next;
    }

    int group, flag, size = temp.size();
    if (size % K == 0) {
        flag = 0;
        group = size / K;
    }
    else {
        flag = 1;
        group = size / K + 1;
    }

    int pre;
    for (int i = 0; i < group; ++i) {
        if (i < group - 1) {
            for (int j = K - 1; j >= 0; j--) {
                if (j > 0)
                    pre = i * K + j - 1;
                else if (j == 0) {
                    if (i < group - 2 )
                        pre = K * (i + 1) + K - 1;
                    else {
                        if (flag)
                            pre = K * (i + 1);
                        else
                            pre = K * (i + 1) + K - 1;
                    }
                }
                printf("%05d %d %05d\n", temp[i * K + j].first, temp[i * K + j].second, temp[pre].first);
            }
        } else {
            if (flag) {
                for (int j = K * (group - 1); j < temp.size() - 1; ++j) {
                    printf("%05d %d %05d\n", temp[j].first, temp[j].second, temp[j + 1].first);
                }
                printf("%05d %d -1\n", temp[temp.size() - 1].first, temp[temp.size() - 1].second);
            } else {
                for (int j = K - 1; j >= 0; j--) {
                    if (j > 0) {
                        pre = (group - 1) * K + j - 1;
                        printf("%05d %d %05d\n", temp[i * K + j].first, temp[i * K + j].second, temp[pre].first);
                    }
                    else if (j == 0) {
                        printf("%05d %d -1\n", temp[(group - 1) * K].first, temp[(group - 1) * K].second);
                    }
                }
            }
        }
    }
    return 0;
}