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 (≤) which is the total number of nodes, and a positive K (≤) 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 = 100010; struct Node{int address,data,next;int flag; }node[maxn];bool cmp(Node a,Node b){return a.flag < b.flag; }//void print(int i,int k,int n){ // int j; // for(j = i + k - 1; j >= i; j--){ // printf("%05d %d ",node[j].address,node[j].data); // if(j > i){ // printf("%05d\n",node[j-1].address); // }else{ // if(i+2*k<=n){ // printf("%05d\n",node[i+2*k].address); // } // } // } //}int main(){for(int i = 0 ; i < maxn; i++){node[i].flag = maxn;}int n,k,begin;scanf("%d%d%d",&begin,&n,&k);int address;for(int i = 0; i < n; i++){scanf("%d",&address);scanf("%d%d",&node[address].data,&node[address].next);node[address].address = address;//node[address].flag = maxn; }int count = 0,p = begin;while(p!=-1){node[p].flag = count++;p = node[p].next;}sort(node,node+maxn,cmp); n = count; for(int i = 0; i < n/k; i++){for(int j = (i+1)*k-1; j > i*k; j--){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);if(i < n/k-1) printf("%05d\n",node[(i+2)*k-1].address);else{if(n%k==0) printf("-1\n");else{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");}}}}// if(n == 1){ // printf("%05d %d -1\n",node[0].address,node[0].data); // return 0; // } // bool flag = true; // if(count%k!=0){ // while(count%k!=0) count--; // flag = false; // } // for(int i = 0; i < count; i += k){ // print(i,k,count); // } // // if(flag == true){ // printf("-1\n"); // }else{ // printf("%05d\n",node[count].address); // for(int i = count; 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"); // } // } // // for(int i = 0; i < n; i++){ // printf("%05d %d\n",node[i].address,node[i].data); // }return 0; }
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