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; }
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