Luogu · P1451

求细胞数量

This C++ solution uses graph traversal for Luogu P1451 求细胞数量. Read the reasoning, inspect the code, or try your own test case below.

P1451Graphs & treesGraph traversalC++45 lines
Solution243of 248
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Approach

Graph traversal

求细胞数量 matches the digit grid and four-directional connected-component count of nonzero cells.

Graphs & trees

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

P_1451_求细胞数量.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int n, m; cin >> n >> m;
        vector<vector<char>> vis(n, vector<char>(m));
        for (int i = 0; i < n; i++) {
            string s; cin >> s;
            for (int j = 0; j < s.size(); j++) {
                if (s[j] == '0') vis[i][j] = 1;
            }
        }
        int cnt = 0;
        vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        auto bfs = [&](int i, int j) {
            queue<pair<int, int>> q;
            q.push({i, j});
            vis[i][j] = 1;
            while (!q.empty()) {
                auto[r, c] = q.front(); q.pop();
                for (int d = 0; d < dir.size(); d++) {
                    int nr = r + dir[d].first, nc = c + dir[d].second;
                    if (nr >= 0 && nr < n && nc >= 0 && nc < m && !vis[nr][nc]) {
                        vis[nr][nc] = 1;
                        q.push({nr, nc});
                    }
                }
            }
            return;
        };
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!vis[i][j]) {
                    bfs(i, j);
                    cnt++;
                }
            }
        }
        cout << cnt << '\n';
        return 0;
    }
        

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