A1052. Linked List Sorting

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

#include<cstdio>
#include<iostream>
#include<algorithm>
typedef struct NODE{
    int lt, rt;
    int data;
    int valid;
    NODE(){
        valid = 0;
    }
}node;
node nds[100001];
bool cmp(node a, node b){
    if(a.valid != b.valid){
        return a.valid > b.valid;
    }else return a.data < b.data;
}
using namespace std;
int main(){
    int N, head, addr;
    scanf("%d%d", &N, &head);
    for(int i = 0; i < N; i++){
        scanf("%d", &addr);
        nds[addr].lt = addr;
        scanf("%d %d", &nds[addr].data, &nds[addr].rt);
    }
    int pt = head, cnt = 0;
    while(pt != -1){
        nds[pt].valid = 1;
        cnt++;
        pt = nds[pt].rt;
    }
    sort(nds, nds + 100001, cmp);
    if(cnt > 0)
        printf("%d %05d\n", cnt, nds[0].lt);
    else printf("0 -1");
    for(int i = 0; i < cnt; i++){
        if(i == cnt - 1){
            printf("%05d %d -1\n", nds[i].lt, nds[i].data);
        }else{
            printf("%05d %d %05d\n", nds[i].lt, nds[i].data, nds[i + 1].lt);
        }
    }
    cin >> N;
    return 0;
}

总结：

1、题意：将所给的节点重新排序。 但注意，仅仅是合法节点进行排序。合法节点是指按照第一行所给的首地址能串在一起的所有节点。题目会给孤立的非法节点，需要注意区分。办法就是对合法节点标记1，非法节点标记0。然后利用排序进行分类，以及二级排序。

2、当没有合法节点时，要注意输出特殊情况 0 -1。

3、在读入每一个节点的时候，不要忘记给节点的地址赋值。