zoukankan      html  css  js  c++  java
  • 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 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
    

     

    需要判断是否为空链,即判断初试地址是否为-1。

    地址没有必要用string和char[],使用string会超时,使用char[]时无法使用map,完全可以使用int,输出时用%05d。

    #include <bits/stdc++.h>
    
    using namespace std;
    
    int N;
    int s;
    struct Node {
        int address;
        int next;
        int key;
        bool operator < (const Node &n) const {
            return key < n.key;
        }
    }nodes[100005], nodesAns[100005];
    map<int, int> nextMap;
    map<int, Node> nodeMap;
    
    int main()
    {
        cin>>N;
        cin>>s;
        for(int i = 0; i < N; i++) {
            scanf("%d%d%d", &nodes[i].address, &nodes[i].key, &nodes[i].next);
            nextMap[nodes[i].address] = nodes[i].next;
            nodeMap[nodes[i].address] = nodes[i];
        }
        if(s == -1) {
            printf("0 -1");
            return 0;
        }
        int index = 0;
        while(1) {
            if(s == -1) break;
            nodesAns[index] = nodeMap[s];
            index++;
            s = nextMap[s];
        }
        sort(nodesAns, nodesAns+index);
        //cout<< index<< " "<< nodesAns[0].address<< endl;
        printf("%d %05d
    ", index, nodesAns[0].address);
        for(int i = 0; i < index; i++) {
            if(i != index-1) printf("%05d %d %05d
    ", nodesAns[i].address, nodesAns[i].key, nodesAns[i+1].address);
            else {
                printf("%05d %d -1", nodesAns[i].address, nodesAns[i].key);
            }
        }
        return 0;
    }
  • 相关阅读:
    Java 多线程系列02
    Java 多线程系列01
    java io流03 字符流
    java JDBC系列
    java io流02 字节流
    Helidon使用心得
    camel 解析
    Spring 源码分析
    java代码实现分页功能
    SpringBoot Tomcat启动报错
  • 原文地址:https://www.cnblogs.com/ACMessi/p/8443467.html
Copyright © 2011-2022 走看看