矩阵染色(Matrix Coloring) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<string> g(n);
    for (int i = 0; i < n; ++i) cin >> g[i];
    vector<int> useC(n, 0), useR(n, 0);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (g[i][j] == 'B') useR[i] = 1;
            else useC[j] = 1;
        }
    }
    int totUsed = 0;
    for (int j = 0; j < n; ++j) totUsed += useC[j];
    for (int i = 0; i < n; ++i) totUsed += useR[i];

    // Nodes 0..n-1: C1..Cn; Nodes n..2n-1: R1..Rn
    vector<vector<int>> adj(2 * n);
    vector<int> deg(2 * n, 0);
    for (int i = 0; i < n; ++i) {
        if (!useR[i]) continue;
        for (int j = 0; j < n; ++j) {
            if (!useC[j]) continue;
            if (g[i][j] == 'B') {
                adj[j].push_back(n + i);
                deg[n + i]++;
            } else {
                adj[n + i].push_back(j);
                deg[j]++;
            }
        }
    }
    priority_queue<int, vector<int>, greater<int>> pq;
    for (int j = 0; j < n; ++j) {
        if (useC[j] && deg[j] == 0) pq.push(j);
    }
    for (int i = 0; i < n; ++i) {
        if (useR[i] && deg[n + i] == 0) pq.push(n + i);
    }
    vector<string> ans;
    while (!pq.empty()) {
        int u = pq.top();
        pq.pop();
        if (u < n) ans.push_back("C" + to_string(u + 1));
        else ans.push_back("R" + to_string(u - n + 1));
        for (int v : adj[u]) {
            if (--deg[v] == 0) pq.push(v);
        }
    }
    if ((int)ans.size() < totUsed) {
        cout << "No solution\n";
    } else {
        for (int i = 0; i < (int)ans.size(); ++i) {
            cout << ans[i] << (i + 1 == (int)ans.size() ? "\n" : " ");
        }
    }
    return 0;
}