1074. Reversing Linked List (25)

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

---

1、本题是栈和队列的结合，我使用了vector代替队列，功能一致。
2、注意case6，有的节点不在链中，所以需要先算出链条长度。

---

#include <iostream>
#include <map>
#include <vector>
#include<stack>
using namespace std;
struct node
{
    int ad,num,next;
    node(){}
    node(int a,int nu,int ne){
        ad=a,num=nu,next=ne;
    }
};
node mnode[100005];
int main(){
    freopen("in.txt","r",stdin);
    int start,n,m;
    stack<node> out;
    vector<node> out2;
    while(scanf("%d%d%d",&start,&n,&m)!=EOF)
    {
        int tad,tnum,tnext;
        for(int i=0;i<n;i++)
        {
            cin>>tad>>tnum>>tnext;
            mnode[tad]=node(tad,tnum,tnext);
        }
    }

    int now=start,i,t=start,cnt=0;
    n=0;
    while(t!=-1){//计算链条长度
        n++;
        t=mnode[t].next;
    }
    for(i=0;i+m<=n;i+=m){
        for(int j=i;j<i+m;j++){
            out.push(mnode[now]);
            now=mnode[now].next;
        }
        while(!out.empty()){
            out2.push_back(out.top());
            out.pop();
        }
    }
    for(;i<n;i++){
        out2.push_back(mnode[now]);
        now=mnode[now].next;
    }
    for(int i=0;i<out2.size()-1;i++){
        printf("%05d %d %05d\n",out2[i].ad,out2[i].num,out2[i+1].ad);
    }
    printf("%05d %d -1\n",out2[n-1].ad,out2[n-1].num);
    return 0;
}