pat 1052
2014-03-08 10:44
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1052. Linked List Sorting (25)
时间限制 400 ms内存限制 32000 kB代码长度限制 16000 B判题程序 StandardA linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and aNext pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.Input Specification:Each input file contains one test case. For each case, the first line contains a positive N (< 105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by -1.Then N lines follow, each describes a node in the format:Address Key Nextwhere Address is the address of the node in memory, Key is an integer in [-105, 105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.Output Specification:For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.Sample Input:5 00001 11111 100 -1 00001 0 22222 33333 100000 11111 12345 -1 33333 22222 1000 12345Sample Output:
5 12345 12345 -1 00001 00001 0 11111 11111 100 22222 22222 1000 33333 33333 100000 -1
#include <stdio.h>#include <algorithm>using namespace std;struct Node{ int pre; int data; int next;} node[100010], nn[100010];bool cmp(Node a, Node b){ return a.data < b.data;}int main(){ int n, head, pre, data, next; int i, j; scanf("%d%d", &n, &head); if(head == -1) {//如果链表为空,输出0,-1 printf("0 -1\n"); return 0; } for(i = 0; i < n; i++) { scanf("%d%d%d", &pre, &data, &next); node[pre].pre = pre; node[pre].data = data; node[pre].next = next; } //将不是链表中的结点删除 j = head; i = 0; while(j != -1) { nn[i].pre = node[j].pre; nn[i].data = node[j].data; nn[i].next = node[j].next; j = node[j].next; i++; } n = i; sort(nn, nn + i, cmp); head = nn[0].pre; printf("%d %05d\n", n, head); for(i = 0; i < n; i++) { if(i != n-1) printf("%05d %d %05d\n", nn[i].pre, nn[i].data, nn[i+1].pre); else printf("%05d %d -1\n", nn[i].pre, nn[i].data); } return 0;}[/code]
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