Pat 1052 Linked List Sorting (25)

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 0000111111 100 -100001 0 2222233333 100000 1111112345 -1 3333322222 1000 12345

Sample Output:

5 1234512345 -1 0000100001 0 1111111111 100 2222222222 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 100000struct 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;}