PatA 1038 Recover the Smallest Number-拼接最小数 2020/8/13

问题描述

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:
Each input file contains one test case. Each case gives a positive integer N (≤10​⁴) followed by N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.

Output Specification:
For each test case, print the smallest number in one line. Notice that the first digit must not be zero.

解决方法

#include 
#include 
#include 
#include 
using namespace std;
vector v;
//判断字符串有多少前导零
int calcZero(string s)
{
    int n = 0;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] != '0')
            break;
        n++;
    }
    return n;
}
//核心比较函数
bool cmp(string s1, string s2)
{
    int pre1 = calcZero(s1);
    int pre2 = calcZero(s2);
    //前导0多的放前面
    if (pre1 != pre2)
    {
        return pre1 > pre2;
    }
    else
    {
        int len1 = s1.length();
        int len2 = s2.length();
        //长度相同的比较大小
        if (len1 == len2)
        {
            return s1 < s2;
        }
        /* 长度不相等的 先看短的字符串和长的相同长度的前缀比较大小,要小的在前。
            如果比较结果相等就通过拼接比较来判断那个应该排在前。
        */
        else if (len1 > len2)
        {
            //相同长度的前缀比较
            if (s1.substr(0, len2) != s2)
            {
                return s1.substr(0, len2) < s2;
            }
            //拼接比较先后顺序
            else
            {
                return s1.substr(len2) + s2 < s1;
            }
        }
        else
        {
            //相同长度的前缀比较
            if (s2.substr(0, len1) != s1)
            {
                return s2.substr(0, len1) > s1;
            }
            //拼接比较先后顺序
            else
            {
                return s2.substr(len1) + s1 > s2;
            }
        }
    }
}


int main(void)
{
    //输入数据并排序
    int m;
    string str = "";
    scanf("%d", &m);
    for (int i = 0; i < m; i++)
    {
        string s;
        cin >> s;
        v.push_back(s);
    }
    sort(v.begin(), v.end(), cmp);

    //根据排序结果 拼接字符串str
    for (vector::iterator it = v.begin(); it != v.end(); it++)
    {
        str += *it;
    }

    //清除结果的前导0,如果导致结果为空则要输出0 测试用例2就是这样的情形
    int begin = calcZero(str);
    str = str.substr(begin);
    if (str.length() == 0)
    {
        cout << 0 << endl;
    }
    else
    {
        cout << str << endl;
    }
    return 0;
}

基本策略(cmp)

• 先按前导零个数选择顺序尽量减少位数和将0放在高位
• ,前导零个数相同的长度相等的按字典序比较小的放前面
• 长度不同的如s1>s2,先从s1前缀中截取s2长度的字符串比较小的放前面
• 如果比较还相等就将s1剩余部分+s2来和s1本身比较确定顺序。(s1s2,s2s1)

遇到的问题

• 本来挺有自信的但是测试用例2总报错,才发现当拼接出的字符串全为0时结果会为空

结果展示

PatA1038.png