1074 Reversing Linked List (25)（25 分）
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 (<= 10^5^) 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 33218Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1这题刚开始用list和map写 超时了
后来改成数组的写法
嗯.......最后一个测试点错了 我也不知道什么原因 是看网上的 他人的博客说是 可能存在！！！！！那种 不需要在这个链表出现的结点。。。嗯 改了下就对了
代码:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 1e5+100;
int a[maxn],c[maxn];
int b[maxn][2];
int main()
{
int digit,n,m;
scanf("%d%d%d",&digit,&n,&m);
for(int i = 1; i <= n; i++)
{
int x,y,z;
scanf("%d%d%d",&x, &y, &z);
b[x][0] = y;
b[x][1] = z;
}
int gg = 0;
a[++gg] = digit;
for(++gg;a[gg-1] != -1;++gg)
a[gg] = b[a[gg-1]][1];
gg--;gg--;
// cout<<gg<<endl;
// for(int i =1 ; i<=gg;i++)
// cout<<a[i]<<endl;
int now = 0,i;
for(i = m;i <= gg; i += m )
{
for(int j = i; j > i-m; j--){
c[++now] = a[j];
}
}
if(i>gg){
for(i = i- m + 1; i <= gg;i++)
c[++now] = a[i];
}
for(int i = 1 ; i < gg;i++)
printf("%05d %d %05d\n",c[i],b[c[i]][0],c[i+1]);
printf("%05d %d %d\n",c[gg],b[c[gg]][0],-1);
return 0;
}