02-1. Reversing Linked List (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
8000 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
这题用二维数组高维代表首地址啊address,低维[address][0]储存data,[address][1]则储存next,牺牲空间换时间。
再使用维数组list进行reverse.
1 int list[100010]; 2 int node[100010][2]; 3 4 int st,num,r; 5 cin>>st>>num>>r; 6 int address,data,next,i; 7 for(i=0;i<num;i++) 8 { 9 cin>>address>>data>>next; 10 node[address][0]=data; 11 node[address][1]=next; 12 }
先输入node值。如图
address | 00000 | 00100 | 12309 | 33218 | 68237 | 99999 |
data | 4 | 1 | 2 | 3 | 6 | 5 |
next | 99999 | 12309 | 33218 | 00000 | -1 | 68237 |
值得说明的是:不需要按照顺序输入,到数组中后自然会对应数组的下标值即address。
因此上表中按照数组顺序排序,无初始值的数组下标省略不列。
下面对list赋address.
int m=0,n=st; while(n!=-1) { list[m++]=n; n=node[n][1]; }
list | 0 | 1 | 2 | 3 | 4 | 5 |
address | 00100 | 12309 | 33218 | 00000 | 99999 | 68237 |
其中list[0]储存的是st的address,list[1]储存的是st->next的地址,以此类推。直到遇到address为-1.
1 i=0; 2 while(i+r<=m) 3 { 4 reverse(list+i,list+i+r); 5 i=i+r; 6 }
进行反转,使用algorithm的reverse函数。
1 for (i = 0; i < m-1; i++) 2 { 3 printf("%05d %d %05d\n", list[i], node[list[i]][0], list[i+1]); 4 } 5 printf("%05d %d -1\n", list[i], node[list[i]][0]);
最后进行输出,注意的是以int形式储存的,如address中的00000实际储存位置是0,故输出时注意是要用五位数输出。
完整AC代码如下:
1 #include<iostream> 2 #include<algorithm> 3 using namespace std; 4 int list[100010]; 5 int node[100010][2]; 6 int main() 7 { 8 freopen("F:\\in1.txt","r",stdin); 9 int st,num,r; 10 cin>>st>>num>>r; 11 int address,data,next,i; 12 for(i=0;i<num;i++) 13 { 14 cin>>address>>data>>next; 15 node[address][0]=data; 16 node[address][1]=next; 17 } 18 int m=0,n=st; 19 while(n!=-1) 20 { 21 list[m++]=n; 22 n=node[n][1]; 23 } 24 i=0; 25 while(i+r<=m) 26 { 27 reverse(list+i,list+i+r); 28 i=i+r; 29 } 30 for (i = 0; i < m-1; i++) 31 { 32 printf("%05d %d %05d\n", list[i], node[list[i]][0], list[i+1]); 33 } 34 printf("%05d %d -1\n", list[i], node[list[i]][0]); 35 }