02-线性结构1. 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
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
struct node{
int now,next,data;
};
node mem[];
vector<node> v;
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
int h,now,data,next,num,k,temp;
scanf("%d %d %d",&h,&num,&k);
int i;
for(i=;i<num;i++){
scanf("%d %d %d",&now,&data,&next);
mem[now].now=now;
mem[now].data=data;
mem[now].next=next;
}
now=h;
while(now!=-){
v.push_back(mem[now]);
now=mem[now].next;
}
int length=v.size();
int round=length/k;
for(i=;i<round;i++){
reverse(v.begin()+i*k,v.begin()+i*k+k);
}
for(i=;i<length-;i++){
printf("%05d %d %05d\n",v[i].now,v[i].data,v[i+].now);
}
printf("%05d %d %d\n",v[i].now,v[i].data,-);//注意全反的情况
return ;
}