GYM - 101502I - Move Between Numbers

Question

题目大意

就是给你 n ， s 、 e ， n 代表的是有多少个字符串，下面每一个字符串可以看做是图上的一个顶点，如果两个字符串之间满足题目所给的那个条件，即: 两个字符串的相同的字符个数的数目 sum = 17 (可以这样理解，就是题目那个等式的意思)，两个字符串（顶点) 就可以相互到达，问从 s 到 e 最少的步数 mindistance 。

解析

编写一个函数判断两个点是否可以相互到达，然后用 dijkstra 模板 跑一下即可。

import java.io.BufferedInputStream;
import java.util.*;

public class Main {

    static int n; // vertex num
    static int m; // edge num
    static boolean[] vis;
    static ArrayList<Edge>[] G;

    static class Edge implements Comparable<Edge> {
        public int to;
        public int w;

        public Edge(int to, int w) {
            this.to = to;
            this.w = w;
        }

        @Override
        public int compareTo(Edge o) {
            return w - o.w;
        }
    }

    static boolean common(String a, String b) {
        int[] ca = new int[10];
        int[] cb = new int[10];
        for (int i = 0; i < a.length(); i++) {
            ca[a.charAt(i) - '0']++;
            cb[b.charAt(i) - '0']++;
        }
        int sum = 0;
        for (int i = 0; i < 10; i++)
            sum += Math.min(ca[i], cb[i]);
        return sum == 17;
    }

    static int[] dijkstra(int start) {
        PriorityQueue<Edge> pq = new PriorityQueue<>();
        int[] dis = new int[n + 1];
        for (int i = 0; i <= n; i++)
            dis[i] = Integer.MAX_VALUE;
        dis[start] = 0;
        // G.vis[str] = true; // 下面 Queue 还要访问 start
        pq.add(new Edge(start, 0)); //自环边
        while (!pq.isEmpty()) {
            Edge curEdge = pq.poll();
            int to = curEdge.to;
            if (vis[to])
                continue;
            vis[to] = true;
            for (int i = 0; i < G[to].size(); i++) {
                int nxtNode = G[to].get(i).to;
                int nxtW = G[to].get(i).w;
                if (!vis[nxtNode] && dis[nxtNode] > dis[to] + nxtW) {
                    dis[nxtNode] = dis[to] + nxtW;
                    pq.add(new Edge(nxtNode, dis[nxtNode]));
                }
            }
        }
        return dis;
    }

    public static void main(String[] args) {

        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        int T = cin.nextInt();
        while (T-- > 0) {
            n = cin.nextInt();
            int s = cin.nextInt();
            int e = cin.nextInt();
            String[] str = new String[n];
            for (int i = 0; i < n; i++)
                str[i] = cin.next();
            G = new ArrayList[n + 1];
            vis = new boolean[n + 1];
            for (int i = 0; i <= n; i++) {
                G[i] = new ArrayList<>();
                vis[i] = false;
            }
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    if (common(str[i], str[j])) {
                        G[i + 1].add(new Edge(j + 1, 1)); // notice i+1, j+1
                        G[j + 1].add(new Edge(i + 1, 1));
                    }
                }
            }
            int[] dis = dijkstra(s);
            System.out.println(dis[e] == Integer.MAX_VALUE ? -1 : dis[e]);
        }
    }
}

C++ 代码:

#include <bits/stdc++.h>

const int maxn = 251;
const int INF = 1e9;

class Edge{
public:
    int to, w;
    Edge(int to, int w):to(to), w(w){}
    // bool operator < (const Edge& rhs) const {
    //     return w > rhs.w;
    // }
};

bool operator < (const Edge& ea, const Edge& eb){ 
    return ea.w > eb.w;
}

std::vector<Edge>G[maxn];
bool vis[maxn];
int dis[maxn];

bool common(char *s1,char *s2){
    int len = strlen(s1);
    int a[10] = {0}, b[10] = {0};
    for(int i = 0; i < len; i++)a[s1[i]-'0']++;
    for(int i = 0; i < len; i++)b[s2[i]-'0']++;
    int sum = 0;
    for(int i = 0; i <= 9; i++)
        sum += std::min(a[i], b[i]);
    return sum == 17;
}

int dijkstra(int s, int e, int n){
    // priority_queue<Type, Container, Functional>
    // priority_queue 使用:  https://blog.csdn.net/bat67/article/details/77585312 
    std::priority_queue<Edge>pq; //比较方式默认用 operator< ,所以俩个参数缺省的话，优先队列就是大顶堆，队头元素最大。
    for(int i = 0; i < n; i++) dis[i] = INF;
    dis[s] = 0;
    pq.push(Edge(s, 0));
    // vis[d] = true;
    while(!pq.empty()){
        Edge cur = pq.top(); pq.pop();
        int to = cur.to;
        if(vis[to])
            continue;
        vis[to] = true;
        for(int i = 0; i < G[to].size(); i++){
            int to2 = G[to][i].to;
            int w = G[to][i].w;
            if(!vis[to2] && dis[to2] > w+dis[to]){
                dis[to2] = w+dis[to];
                pq.push(Edge(to2, dis[to2]));
            }
        }
    }
    return dis[e];
}

int main(){
    int T, n, m, s, e;
    char str[250][25];
    std::cin >> T;
    while(T--){
        std::cin >> n >> s >> e;
        for(int i = 0; i < n; i++)
            std::cin >> str[i];
        for(int i = 0; i < n; i++) G[i].clear();
        for(int i = 0; i < n; i++) vis[i] = false;
        for(int i = 0; i < n; i++){
            for(int j = i+1; j < n; j++){
                if(common(str[i], str[j])){
                    G[i].push_back(Edge(j,1));
                    G[j].push_back(Edge(i,1));
                }
            }
        }
        int res = dijkstra(s-1,e-1, n); // notice index
        std::cout << (res == INF ? -1 : res) << std::endl;
    }
    return 0;
}

题目链接

https://codeforces.com/gym/101502/problem/I