Pat 1052 Linked List Sorting (25)
1052. Linked List Sorting (25)
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 巨坑,有链表是空的情况,还有点不再链表中,还有千万别用cin#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string>
#include <map> using namespace std;
#define MAX 100000
struct Node{
char l[10],r[10];
int key;
}a[MAX+5],e[MAX+5];
int cmp(Node a,Node b){
return a.key<b.key;
}
int n;
char b[10];
map<string,string> m1;
map<string,int> m2;
int main()
{
scanf("%d",&n);scanf("%s",b);
m1.clear();m2.clear();
for(int i=1;i<=n;i++){
scanf("%s%d%s",a[i].l,&a[i].key,a[i].r);
m1[a[i].l]=a[i].r;m2[a[i].l]=i;
}
string c=b;
if(c=="-1"){
cout<<0<<" "<<b<<endl;
return 0;
}
int cnt=0;
while(c!="-1"){
e[++cnt]=a[m2[c]];
c=m1[c];
}
sort(e+1,e+cnt+1,cmp);
cout<<cnt<<" "<<e[1].l<<endl;
for(int i=1;i<=cnt-1;i++)
cout<<e[i].l<<" "<<e[i].key<<" "<<e[i+1].l<<endl;
cout<<e[cnt].l<<" "<<e[cnt].key<<" "<<"-1"<<endl;
return 0;
}
Pat 1052 Linked List Sorting (25)的更多相关文章
- PAT 解题报告 1052. Linked List Sorting (25)
1052. Linked List Sorting (25) A linked list consists of a series of structures, which are not neces ...
- 【PAT】1052 Linked List Sorting (25)(25 分)
1052 Linked List Sorting (25)(25 分) A linked list consists of a series of structures, which are not ...
- PAT 甲级 1052 Linked List Sorting (25 分)(数组模拟链表,没注意到不一定所有节点都在链表里)
1052 Linked List Sorting (25 分) A linked list consists of a series of structures, which are not ne ...
- PAT 1052 Linked List Sorting [一般]
1052 Linked List Sorting (25 分) A linked list consists of a series of structures, which are not nece ...
- PAT甲题题解-1052. Linked List Sorting (25)-排序
三个注意点: 1.给出的n个节点并不一定都在链表中 2.最后一组样例首地址即为-1 3.输出地址的时候一直忘记前面要补0... #include <iostream> #include & ...
- PAT Advanced 1052 Linked List Sorting (25) [链表]
题目 A linked list consists of a series of structures, which are not necessarily adjacent in memory. W ...
- PAT (Advanced Level) 1052. Linked List Sorting (25)
简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...
- 【PAT甲级】1052 Linked List Sorting (25 分)
题意: 输入一个正整数N(<=100000),和一个链表的头结点地址.接着输入N行,每行包括一个结点的地址,结点存放的值(-1e5~1e5),指向下一个结点的地址.地址由五位包含前导零的正整数组 ...
- 1052. Linked List Sorting (25)
题目如下: A linked list consists of a series of structures, which are not necessarily adjacent in memory ...
随机推荐
- vue TypeError: Cannot read property 'length' of undefined 问题解决思路
上图中是我在接手一个项目的时候,出现的一个关于数组的 bug,但是从上图来看,其实报错是不清楚的,只能从报错信息上看出来是 /src/page/waybill/waybill_form.vue 这个 ...
- 无法删除尚未附加的实体-linq中使用反射删除记录出现错误,应该如何解决。
db.airports.Attach(data); db.airports.DeleteOnSubmit(data); db.SubmitChanges();
- ContentObserver与DatasetObserver区别
1. ContentObserver ContentObserver主要是通过Uri来监测特定的Databases的表,如果该Databases表有变动则会通知更新cursor中的数据. 如果 ...
- css 禁用移动端部分特性
禁止复制黏贴功能 .oncopy{ -webkit-user-select: none; } 禁止点击之后修改背景颜色 -webkit-tap-highlight-color: rgba(0,0,0, ...
- mysql init_connect
init_connect 服务器为每个连接的客户端执行的字符串.字符串由一个或多个SQL语句组成.要想指定多个语句,用分号间隔开.例如,每个客户端开始时默认启用autocommit模式.没有全局服务器 ...
- oblique perspective projection
参考: https://en.wikibooks.org/wiki/GLSL_Programming/Vertex_Transformations <3D游戏与计算机图形学中的数学方法>E ...
- MONGODB Date 处理方法
mongodb 日期处理:1,用new Date()存入数据库,要转一下.2,输出的显示的时候,要把data 后的Z 去啦.3, 查询时数据时不用处理.
- 点滴积累【JS】---JS小功能(JS实现动态添加运动属性)
效果: 思路: 首先遍历div挨个执行onmouseover事件,再设置获取非行间样式.然后编写setInterval计时器框架,框架内容是:将三个参数 div.div属性.div的目标点,分别获得, ...
- 天猫 小游戏 24 point
游戏规则:给你四个整数,当然他给的是有解的,然后用' + - * / ( ) ,这几种符号任意组合,使运算结果等于24; 用代码快速解决问题,呵呵... #include<io ...
- HDU 1867 A + B for you again(KMP算法的应用)
A + B for you again Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Othe ...