pat-a1074. Reversing Linked List (25)

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之前做过，代码比以前精简一点了。。虽然偷懒用了map..

测试点其实也不严谨，只有最后一个测试点是没有用完给出的所有节点。。

后面要做点ccf的题了。。希望今年能取得好成绩

#include<cstdio>#include<cstring>#include<map>using namespace std;int num[100010];int nu[100010];int main(){int a,b,c,x,y,z;scanf("%d%d%d",&a,&b,&c);map<int,int> an;map<int,int> na;map<int,int> nl;for(int i=0;i<b;++i){scanf("%d%d%d",&x,&y,&z);an[x]=y;na[y]=x;nl[y]=z;}int t=0;while(a!=-1){num[t++]=an[a];a=nl[an[a]];}int p=t/c;int k=0;for(int i=1;i<=p;++i){for(int j=i*c-1;j>=(i-1)*c;--j)  nu[k++]=num[j];}for(int i=c*p;i<t;++i) nu[k++]=num[i];for(int i=0;i<t-1;++i) printf("%05d %d %05d\n",na[nu[i]],nu[i],na[nu[i+1]]);printf("%05d %d -1\n",na[nu[t-1]],nu[t-1]);return 0;}

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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

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