제출 #1290841

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1290841		ecen30	세계 지도 (IOI25_worldmap)	C++20	0 / 100	2 ms	572 KiB

worldmap

//testing AI COde
#include "worldmap.h"
#include <bits/stdc++.h>
using namespace std;

// We create a DFS order of the adjacency graph and use it as the linear ordering.
// Then we place that ordering onto a cyclic Latin torus grid of size K = N.

vector<vector<int>> create_map(int N, int M, vector<int> A, vector<int> B) {
    vector<vector<int>> G(N+1);
    for (int i = 0; i < M; i++) {
        G[A[i]].push_back(B[i]);
        G[B[i]].push_back(A[i]);
    }

    // --- Step 1: DFS order to linearize the adjacency graph ---
    vector<int> order;
    vector<int> vis(N+1, 0);

    function<void(int)> dfs = [&](int u) {
        vis[u] = 1;
        order.push_back(u);
        for (int v : G[u]) if (!vis[v]) dfs(v);
    };

    // Graph might be disconnected → start DFS from all nodes.
    for (int i = 1; i <= N; i++)
        if (!vis[i]) dfs(i);

    // Build reverse mapping: newLabel[country] = position in DFS order
    vector<int> newLabel(N+1);
    for (int i = 0; i < N; i++)
        newLabel[order[i]] = i + 1;  // 1..N

    // --- Step 2: K = N ---
    int K = N;
    vector<vector<int>> C(K, vector<int>(K));

    // --- Step 3: Fill torus grid with cyclic pattern ---
    for (int r = 0; r < K; r++) {
        for (int c = 0; c < K; c++) {
            int id = 1 + (r + c) % N;      // id in 1..N
            // Map id back to original country
            int country = order[id - 1];   // inverse of newLabel
            C[r][c] = country;
        }
    }

    return C;
}

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