DMOJ · cco14p1

Troyangles

This C++ solution uses dynamic programming for DMOJ cco14p1 Troyangles. Read the reasoning, inspect the code, or try your own test case below.

cco14p1Dynamic programmingC++37 lines
Solution230of 248
Open official problem ↗ Download C++ file ↓ Search the library → Open full Code Lab ↗ Report an issue ↗

Approach

Dynamic programming

Troyangles matches the triangular grid and the dynamic-programming count of all upright triangles.

Dynamic programming

Problem and code

Useful links.

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

Open official problem ↗View exact source file ↗
Implementation

P_1_Troyangles.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; cin >> n;
        vector<vector<char>> v(n, vector<char>(n));
        vector<vector<int>> dp(n, vector<int>(n));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> v[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            if (v[n - 1][i] == '#') dp[n - 1][i] = 1;
        }
        for (int i = n - 2; i >= 0; i--) {
            for (int j = 0; j < n; j++) {
                char c = v[i][j];
                if (c == '#') {
                    dp[i][j] = 1;
                    if (j - 1 >= 0 && j + 1 < n) dp[i][j] = min({dp[i + 1][j], dp[i + 1][j - 1], dp[i + 1][j + 1]}) + 1;
                }
            }
        }
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dp[i][j]) cnt += dp[i][j];
            }
        }
        cout << cnt << '\n';
        return 0;
    }
        

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗Keep studying →

Test this problem

Run your code here.

Paste your code, run a test case, compare the output, or trace selected values.

Full trace, comparison & stress testing ↗
StatusReady
Output
No run yet.
Diagnostics
No diagnostics yet.

Each run is isolated and has strict limits. Passing one test does not guarantee the judge will accept the solution.