PAT 1041 Linked List Sorting (25)(链表排序)
2016-08-30 09:00
471 查看
题目
1052. Linked List Sorting (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A 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 a Next 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 Next where 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 12345 Sample Output: 5 12345 12345 -1 00001 00001 0 11111 11111 100 22222 22222 1000 33333 33333 100000 -1
解题思路
1.先根据头地址保存这个链表所有有效的队列,然后在排序就行了。2.输出的时候要注意保存的链表为空的情况。
代码
#include<iostream> #include<cstdio> #include<algorithm> #include<vector> using namespace std; int nx[99999],va[99999]; struct node{ int ad,v; node(){} node(int _ad,int _v){ this->ad = _ad; this->v = _v; } }; bool cmp(const node &a, const node &b){ return a.v<b.v; } int main(int argc, char *argv[]) { int n,f; scanf("%d %d",&n,&f); vector<node> b; int tem_a,tem_b,tem_c; for (int i = 0; i < n; ++i) { scanf("%d %d %d",&tem_a,&tem_b,&tem_c); nx[tem_a] = tem_c; va[tem_a] = tem_b; } //将有效的链表存好 int now = f; while (now != -1) { b.push_back(node(now,va[now])); now = nx[now]; } //输出 if (!b.empty()) { sort(b.begin(),b.end(),cmp); printf("%d %05d\n",b.size(),b[0].ad); for (int i = 0; i < b.size()-1; ++i) { printf("%05d %d %05d\n",b[i].ad,b[i].v,b[i+1].ad); } printf("%05d %d %d\n",b[b.size()-1].ad,b[b.size()-1].v,-1); }else { printf("%d %d\n",0,-1); } return 0; }
相关文章推荐
- PAT (Advanced Level) Practise 1052 Linked List Sorting (25)
- PAT (Advanced Level) Practise 1052 Linked List Sorting (25)
- PAT (Advanced Level) 1097. Deduplication on a Linked List (25) 链表去重
- pat甲级1074. Reversing Linked List (25)、乙级1025. 反转链表 (25)
- PAT甲级题解-1097. Deduplication on a Linked List (25)-链表的删除操作
- 1074. Reversing Linked List (25)【链表翻转】——PAT (Advanced Level) Practise
- PAT1052 Linked List Sorting (25)
- PAT-Linked List Sorting (25)
- PAT 1074. Reversing Linked List (25)(链表反转)
- PAT - 甲级 - 1074. Reversing Linked List (25)(链表)
- PAT 1097. Deduplication on a Linked List (25) 剔除链表中的重复节点,set用法,尾插法
- PAT - 甲级 - 1097. Deduplication on a Linked List (25)(链表)
- PAT甲题题解-1074. Reversing Linked List (25)-求反向链表
- PAT 1133. Splitting A Linked List (25) 链表的拆分和合并
- 1025. 反转链表 (25)PAT乙级&&1074. Reversing Linked List (25)PAT甲级
- Pat 1052 Linked List Sorting (25)
- PAT (Advanced Level) 1074. Reversing Linked List (25) 翻转链表
- PAT 1097. Deduplication on a Linked List (25)(链表问题)(链表分段)
- Pat 1052 Linked List Sorting (25)
- PAT 1074. Reversing Linked List (25)