1074. Reversing Linked List (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
此题主要是注意逆序点
我用的方法有点low，注意点如下，当i+k+k<=v.size()时便有下一次，若没有则输出i+k时的地址信息即可
代码如下：
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct Node{
int current;
int data;
int next;
}node[100000];
vector<Node> v;
int main(){
int first,n,k;
scanf("%d%d%d",&first,&n,&k);
int current,data,next;
for(int i=0;i<n;++i){
scanf("%d %d %d",¤t,&data,&next);
node[current].data=data;node[current].next=next;
node[current].current=current;
}
int index=first;
while(index!=-1){
v.push_back(node[index]);
index=node[index].next;
}
int i=0;
for(;i+k<=v.size();i+=k){
if(i+k!=v.size()){
for(int j=i+k-1;j>=i;--j){
if(j!=i){
printf("%05d %d %05d\n",v[j].current,v[j].data,v[j-1].current);
}else{
if(i+k+k<=v.size()){
printf("%05d %d %05d\n",v[j].current,v[j].data,v[i+2*k-1].current);
}else{
printf("%05d %d %05d\n",v[j].current,v[j].data,v[i+k].current);
}
}
}
}else{
for(int j=i+k-1;j>=i;--j){
if(j!=i){
printf("%05d %d %05d\n",v[j].current,v[j].data,v[j-1].current);
}else{
printf("%05d %d -1\n",v[j].current,v[j].data);
}
}
}
}
for(;i<v.size();++i){
if(i!=v.size()-1){
printf("%05d %d %05d\n",v[i].current,v[i].data,v[i+1].current);
}else{
printf("%05d %d -1\n",v[i].current,v[i].data);
}
}
return 0;
}
