算法问题实战策略 CHILDRENDAY

地址 https://algospot.com/judge/problem/read/CHILDRENDAY

题解

 

ac代码

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>

using namespace std;


//当前顶点的序号为here时
//经过序号为edge的边线能到达的顶点序号是多少
int append(int here, int edge, int mod)
{
    int there = here * 10 + edge;
    if (there >= mod) return mod + there % mod;
    return there % mod;
}

//找出仅由digits内的数组组成
//且满足C mod n == m的最小的C
string gifts(string digits, int n, int m)
{
    //如果按照升序排列边线序号
    //就能找出按照字典顺序排在最前面的路径
    sort(digits.begin(),digits.end());

    //白色顶点i表示为0号~n-1号的顶点 灰色顶点i表示为n号~2n-1号的顶点
    //parent[i]=BFS生成树中顶点i的父节点
    //choice[i]=从parent[i]连向i的边线序号
    vector<int> parent(2 * n, -1), choice(2 * n, -1);
    queue<int> q;
    //将白色0号加入队列
    parent[0] = 0;
    q.push(0);
    while (!q.empty()) {
        int here = q.front();
        q.pop();
        for (int i = 0; i < digits.size(); i++) {
            //跟踪边线digits[i[-'0'
            int there = append(here, digits[i] - '0', n);
            if (parent[there] == -1) {
                parent[there] = here;
                choice[there] = digits[i] - '0';
                q.push(there);
            }
        }
    }

    //未能达到灰色m号 则失败！
    if (parent[n + m] == -1) return "IMPOSSIBLE";
    //跟踪连向父节点的路径并计算C
    string ret;
    int here = n + m;
    while (parent[here]  != here) {
        ret += char('0' + choice[here]);
        here = parent[here];
    }
    reverse(ret.begin(), ret.end());
    return ret;
}


int main()
{
    int n;
    cin >> n;
    while (n--) {
        string s;
        int a, b;
        cin >> s >> a >> b;

        cout << gifts(s, a, b) << endl;
    }
}