PAT(链表)——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

题目大意：

反转字符串，每k个元素反转一次。

题目解析：

简单模拟。

具体代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
#define MAXN 100005
struct node{
	int address;
	int data;
	int count;
}v[MAXN];
map<int,node> m;
bool cmp(node a,node b){
	return a.count>b.count;
}
int main()
{
	int start,n,k;
	scanf("%d%d%d",&start,&n,&k);
	for(int i=0;i<n;i++){
		int pre,data,post;
		scanf("%d%d%d",&pre,&data,&post);
		m[pre]={post,data,0};
	}
	int tmp=start,count=0;
	while(tmp!=-1){
		v[count]={tmp,m[tmp].data,count};
		tmp=m[tmp].address;
		count++;
	}
	for(int i=0;i+k<=count;i+=k){
		sort(v+i,v+i+k,cmp);
	}
	
	for(int i=0;i<count-1;i++)
		printf("%05d %d %05d\n",v[i].address,v[i].data,v[i+1].address);
	printf("%05d %d -1\n",v[count-1].address,v[count-1].data);
    return 0;
}