1074 Reversing Linked List (25point(s)) Easy only once *注意链表中的无效节点的特殊情况

基本思想：

利用栈进行压栈，重新调整顺序。单纯利用栈输出的话无法快捷预知后续的节点首地址，计算索引非常麻烦；

关键点：

注意无效节点的统计；

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<vector> 
#include<string>
#include<math.h>
#include<algorithm>
#include<cstring>
#include<map>
#include<queue>
#include<set>
#include<stack>
using namespace std;


const int maxn = 100000;
int n,m,st;

struct node {
    int st;
    int data;
    int ed;
};

node vec[maxn];

vector<node>ls;

int main() {
    stack<node>stk;
    cin >> st >> n >> m;
    int s;
    for (int i = 0; i < n; i++) {
        cin >> s;
        vec[s].st = s;
        cin >> vec[s].data >> vec[s].ed;
    }
    n = 0;
    int temp = st;
    while (temp != -1) {
        n++;
        temp = vec[temp].ed;
    }
    while (n >= m) {
        //当剩余数目足够时
        for (int i = 0; i < m; i++) {
            //进行压栈操作；
            stk.push(vec[st]);
            st = vec[st].ed;
        }
        while (!stk.empty()) {
            node k = stk.top();
            stk.pop();
            n--;
            ls.push_back(k);
        }
    }
    if (n != 0) {
        while (st != -1) {
            ls.push_back(vec[st]);
            st = vec[st].ed;
        }
    }
    for (int i = 0; i < ls.size(); i++) {
        if (i != ls.size() - 1) {
            printf("%05d %d %05d\n", ls[i].st, ls[i].data, ls[i + 1].st);
        }
        else {
            printf("%05d %d -1\n", ls[i].st, ls[i].data);
        }
    }
    return 0;
}