PAT K block链表反转

满分

/**
题意: 每k位反转链表


*/
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
const int N = 100010;
int p[N], w[N];

int main()
{
    int m, n, k;
    cin >> m >> n >> k;
    while (n --)
    {
        int a, b, c;
        scanf("%d %d %d", &a, &c, &b);
        p[a] = b;
        w[a] = c;
    }
    int cnt = 0;
    vector<int> res;
    for(int i = m; i != -1; i = p[i])
    {
        res.push_back(i);
    }
    int size = res.size() % k; // 这一步必须思考清晰,取余则是多余数据
    // size / k 则是你有几个block
    reverse(res.begin(), res.end());
    bool flag = true;
    for(int i = 0; i < res.size(); )
    {
        if (flag) reverse(res.begin() + i, res.begin() + i + size), i += size, flag = false;
        else {
            reverse(res.begin() + i, res.begin() + i + k);
            i += k;
        }
    }
    for(int i = 0; i < res.size(); i++)
    {
        printf("%05d %d ", res[i], w[res[i]]);
        if (i != res.size() - 1) printf("%05d\n", res[i + 1]);
        else printf("-1\n");
    }
}

满分

#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 100010;
int h[N], w[N];

int main()
{
    int s, n, k;
    cin >> s >> n >> k;
    while (n --)
    {
        int a, b, c;
        cin >> a >> c >> b;
        h[a] = b;
        w[a] = c;
    }
    vector<int> res;
    for(int i = s; i != - 1; i = h[i])
    {
        res.push_back(i);
    }
    reverse(res.begin(), res.end());
    int last = res.size() % k;

    int len = res.size() / k;
    reverse(res.begin(), res.begin() + last);

    int pos = last;
    while (len --)
    {
        reverse(res.begin() + pos, res.begin() + pos + k);
        pos += k;
    }
    for(int i = 0; i <res.size(); i++)
    {
        printf("%05d %d ", res[i], w[res[i]]);
        if (i == res.size() - 1) printf("-1\n");
        else printf("%05d\n", res[i + 1]);
    }
}