算法训练营day67

题目1：

#include <iostream>
#include <vector>
#include <string>
#include <unordered_set>
#include <unordered_map>
#include <queue>

using namespace std;

int main() {
    string beginStr, endStr;
    int n;
    cin >> n;
    cin >> beginStr >> endStr;
    unordered_set<string> set;
    for(int i = 0;i < n;i++) {
        string str;
        cin >> str;
        set.insert(str);
    }
    unordered_map<string, int> visitedmp;
    visitedmp.insert(pair<string, int>(beginStr, 1));
    queue<string> qu;
    qu.push(beginStr);
    while(!qu.empty()) {
        string word = qu.front();
        qu.pop();
        int path = visitedmp[word];
        for(int i = 0;i < word.size();i++) {
            string newword = word;
            for(int j = 0;j < 26;j++) {
                newword[i] = j + 'a';
                if(newword == endStr) {
                    cout << path + 1 << endl;
                    return 0;
                }
                if(visitedmp.find(newword) == visitedmp.end() && 
                    set.find(newword) != set.end()) {
                        visitedmp.insert(pair<string, int>(newword, path + 1));
                        qu.push(newword);
                }
            }
        }
    }
    cout << 0 << endl;
    return 0;
}

题目2：105. 有向图的完全可达性 (kamacoder.com)

#include<bits/stdc++.h>

using namespace std;

int n, m;

void dfs(vector<vector<int>>& map, int key, vector<bool>& visited) {
    if(visited[key]) {
        return;
    }
    visited[key] = true;
    for(int i = 1;i <= n;i++) {
        if(map[key][i] == 1) {
            dfs(map, i, visited);
        }
    }
}


int main() {

    cin >> n >> m;
    vector<vector<int>> map(n + 1, vector<int>(n + 1, 0));
    while(m--) {
        int i,j;
        cin >> i >> j;
        map[i][j] = 1;
    }
    
    vector<bool> visited(n + 1, false);
    dfs(map, 1, visited);
    for(int i = 1;i <= n;i++) {
        if(!visited[i]) {
            cout << -1 << endl;
            return 0;
        }
    }
    cout << 1 << endl;
    return 0;
}

题目3：106. 岛屿的周长 (kamacoder.com)

// 其实可以不用dfs因为题目说了中间岛屿是联通的
#include<iostream>
#include<vector>

using namespace std;
int dir[4][2] = {0, -1, -1, 0, 0, 1, 1, 0};
int n, m;
vector<pair<int, int>> result;
int count = 0;

void dfs(vector<vector<int>>& map, vector<vector<bool>>& visited, int x, int y) {
    if(map[x][y] == 0 || visited[x][y]) {
        return;
    }
    result.push_back(pair<int, int>(x, y));
    visited[x][y] = true;
    for(int i = 0;i < 4;i++) {
        int nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        if(nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue;
        dfs(map, visited, nextx, nexty);
    }
}

int main() {
    cin >> n >> m;
    vector<vector<int>> map(n, vector<int>(m));
    for(int i = 0;i < n;i++) {
        for(int j = 0;j < m;j++) {
            cin >> map[i][j];
        }
    }
    vector<vector<bool>> visited(n, vector<bool>(m, false));
    for(int i = 0;i < n;i++) {
        for(int j= 0;j < m;j++) {
            if(map[i][j] != 0 && !visited[i][j]) {
                dfs(map, visited, i, j);
            }
        }
    }
    for(int i = 0;i < result.size();i++) {
        int x = result[i].first;
        int y = result[i].second;
        for(int j = 0;j < 4;j++) {
            int nextx = x + dir[j][0];
            int nexty = y + dir[j][1];
            if(nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) 
            {
                if(nextx == -1 || nextx == n || nexty == -1 || nexty == m) count++;
                continue;
            }
            if(map[nextx][nexty] == 0) count++;
        }
    }
    cout << count << endl;
    return 0;
}