PTA乙级题 1025. 反转链表 (25)

【题目链接】

这题有点难，感觉自己写的代码很差，直接给大佬写的代码吧。

#include<iostream> 
#include<vector> 
#include<algorithm> 
#include<iterator> 
#include<utility> 
using namespace std;  
typedef pair<int, int> P;  
vector<P> coll(100000);  
struct node  
{  
    int adress;  
    int data;  
    int next;  
    node(int a, int b, int c) :adress(a), data(b), next(c){};  
};  
vector<node> vec;  
int main()  
{  
#ifdef ONLINE_JUDGE 
#else 
    freopen("D:\\in.txt", "r", stdin);  
    freopen("D:\\out.txt", "w", stdout);  
#endif 
    int adress, data, next;  
    int n, k, first;  
    while (scanf("%d%d%d", &first, &n, &k) != EOF)  
    {  
        int num = n;  
        while (n--)  
        {  
            scanf("%d%d%d", &adress, &data, &next);  
            P p;  
            p.first = data;  
            p.second = next;  
            coll[adress] = p;  
        }  
        int index = first;  
        int cnt(0);//在链表上的结点个数  
        while (index != -1)  
        {  
            cnt++;  
            vec.push_back(node(index, coll[index].first, coll[index].second));  
            index = coll[index].second;  
        }  
        int t = cnt / k;  
        for (int i = 0; i < t; i++)  
            std::reverse(vec.begin() + i*k, vec.begin() + (i + 1)*k);  
        for (int i = 0; i < vec.size()-1;i++)  
        {  
            printf("%05d %d %05d\n", vec[i].adress, vec[i].data, vec[i + 1].adress);  
        }  
        printf("%05d %d %d\n", vec[vec.size() - 1].adress, vec[vec.size() - 1].data, -1);  

    }  
    return 0;  
}