PAT (Advanced Level) Practise 1097 Deduplication on a Linked List (25)
2016-03-25 21:12
447 查看
1097. Deduplication on a Linked List (25)
时间限制300 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean
time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (<= 105) which is the total number of nodes. 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 Key Next
where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.
Output Specification:
For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 5 99999 -7 87654 23854 -15 00000 87654 15 -1 00000 -15 99999 00100 21 23854
Sample Output:
00100 21 23854 23854 -15 99999 99999 -7 -1 00000 -15 8765487654 15 -1
去重,把去掉的部分放到另外一个链表里。
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn =1e5 + 10;
int n,s,f[maxn],nt[maxn],key[maxn],x,y,z,a,b,aa,bb;
int main()
{
scanf("%d%d", &s,&n);
while (n--)
{
scanf("%d%d%d",&x,&y,&z);
key[x]=y; nt[x]=z;
}
a=b=-1; aa=bb=-1;
for (int i=s;i!=-1;i=nt[i])
{
if (f[abs(key[i])])
{
if (b==-1) b=bb=i;
else bb=nt[bb]=i;
}
else
{
f[abs(key[i])]=1;
if (a==-1) a=aa=i;
else aa=nt[aa]=i;
}
}
nt[aa]=-1;
for (int i=a;i!=-1;i=nt[i])
{
printf("%05d %d ",i,key[i]);
if (nt[i]==-1) printf("-1\n");
else printf("%05d\n",nt[i]);
}
if (b!=-1) nt[bb]=-1;
for (int i=b;i!=-1;i=nt[i])
{
printf("%05d %d ",i,key[i]);
if (nt[i]==-1) printf("-1\n");
else printf("%05d\n",nt[i]);
}
return 0;
}
相关文章推荐
- libdvbpsi源码分析(四)PAT表解析/重建
- PAT配置
- 什么是端口复用动态地址转换(PAT) 介绍配置实例
- MikroTik layer7-protocol
- PAT是如何工作的
- PAT 乙级题:1002. 写出这个数 (20)
- PAT (Advanced Level) Practise 1001-1010
- 数据结构学习与实验指导(一)
- PAT Basic Level 1001-1010解题报告
- 1001. 害死人不偿命的(3n+1)猜想
- 1002. 写出这个数
- 1032. 挖掘机技术哪家强
- 1001. 害死人不偿命的(3n+1)猜想 (PAT basic)
- 1002. 写出这个数(PAT Basic)
- 1004. 成绩排名(PAT Basic)
- 1006. 换个格式输出整数(PAT Basic)
- 1007. 素数对猜想(PAT Basic)
- 1008. 数组元素循环右移问题
- 1009. 说反话(PAT Basic)
- 1011. A+B和C(PAT Basic)