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
#include<cstdio> #include<algorithm> using namespace std; const int maxn = 100100; struct Node{ int address,data,next; int order; }node[maxn]; bool cmp(Node a,Node b){ return a.order < b.order; } int main(){ int i,address; for(i = 0; i < maxn; i++){ node[i].order = maxn; } int begin,n,k; //起始节点地址，节点数目，分组数 scanf("%d%d%d",&begin,&n,&k); for(i = 0; i < n; i++){ scanf("%d",&address); scanf("%d%d",&node[address].data,&node[address].next); node[address].address = address; } int p = begin,count = 0; while(p != -1){ node[p].order = count++; p = node[p].next; } sort(node,node+maxn,cmp); n = count; //因为count=0占一个有效节点，退出循环时，count值就是有效节点 for(i = 0; i < n/k; i++){ //枚举完整的n/k块 for(int j = (i+1)*k - 1; j > i*k; j-- ){ //每块的第i个倒着输出，剩余最后一个节点 printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address); } printf("%05d %d ",node[i*k].address,node[i*k].data); //每块的最后一个节点的前两项数据 if(i < n/k -1) { //如果是非最后一块节点 printf("%05d\n",node[(i+2)*k-1].address); }else{ //如果是最后一块节点 if(n % k == 0) printf("-1\n"); //刚好除整 else{ //如果最后一个节点不规则 printf("%05d\n",node[(i+1)*k].address); for(i = n/k*k; i < n; i++){ printf("%05d %d ",node[i].address,node[i].data); if(i < n - 1){ printf("%05d\n",node[i+1].address); }else{ printf("-1\n"); }// else } // for(i) }//else }//else } //for(i) return 0; }