习题7-10(uva-11214)

最新推荐文章于 2023-11-14 07:00:00 发布

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本文探讨了一个关于棋盘覆盖的经典算法问题。通过深度优先搜索（DFS）策略，该程序尝试寻找在给定的矩阵中放置指定数量标记的最大可能数量。涉及到的数据结构包括一维和多维数组用于记录状态，并通过递归调用实现解决方案。

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int Row = 11, Col = 11;
int n, m, ans;
char matrix[Row][Col];
int vis[4][24];

bool IsOver()
{
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (matrix[i][j] == '.') continue;
			//所在行 列 主对角线  副对角线
			if (!vis[0][i] && !vis[1][j] && !vis[2][j - i + m] && !vis[3][j + i]) return false;
		}
	}
	return true;
}


bool dfs(int depth,int x,int y) {
	if (IsOver()) return true;
	if (depth == 0) return false;
	

	for (int i = x; i < n; i++) {
		for (int j = (i == x)?y:0; j < m; j++) {
			int a = vis[0][i], b = vis[1][j], c = vis[2][j - i + m], d = vis[3][j + i];
			if(a && b && c && d) continue;
			int nx = i + (j + 1) / m;
			int ny = (j + 1) % m;
			
			vis[0][i] = vis[1][j] = vis[2][j - i + m] = vis[3][j + i] = 1;
			if (dfs(depth - 1, nx, ny)) return true;
			vis[0][i] = a;
			vis[1][j] = b;
			vis[2][j - i + m] = c;
			vis[3][j + i] = d;
		}
	}
	return false;
}

int main()
{
	int kCase = 0;
	while (cin >> n && n) {
		cin >> m;
		memset(matrix, '.', sizeof(matrix));
		memset(vis, 0, sizeof(vis));

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				cin >> matrix[i][j];
			}
		}
		ans = max(n, m);
		for (int i = 0; i < ans; i++) {
			if (dfs(i,0,0)) {
				ans = i;
				break;
			}
		}
		cout << "Case " << ++kCase << ": " << ans << endl;
	}
	return 0;
}