利用分治法解决棋盘覆盖问题

最新推荐文章于 2024-05-28 12:14:19 发布

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#分治法

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高效算法——分治法

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本文探讨如何运用分治法解决棋盘覆盖问题。通过深入解析问题本质，结合编程实现，展示了分治策略在解决此类问题上的高效性和实用性。

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1.问题描述：点击打开链接

2.代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<cassert>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<functional>
using namespace std;

#define me(s) memset(s,0,sizeof(s))
#define pb push_back
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair <int, int> P;




const int N = 1 << 11;
int A[N][N];
int n, a, b;
int idx;

void solve(int r1, int r2, int c1, int c2, int x, int y)
{
	if (r2-r1==1)return;
	int t = idx++;
	int mr = (r1 + r2)/2, mc = (c1 + c2)/2;
	if (x<mr&&y<mc)solve(r1, mr, c1, mc, x, y);
	else solve(r1, mr, r2, mc, mr-1,mc-1), A[mr-1][mc-1] = t;

	if (x<mr&&y >= mc)solve(r1, mr, mc, c2, x, y);
	else solve(r1, mr, mc, c2, mr-1,mc), A[mr-1][mc] = t;

	if (x >= mr&&y<mc)solve(mr, r2, c1, mc, x, y);
	else solve(mr, r2, c1, mc, mr,mc-1), A[mr][mc-1] = t;

	if (x >= mr&&y >= mc)solve(mr, r2, mc, c2, x, y);
	else solve(mr, r2, mc, c2, mr,mc), A[mr][mc] = t;
}

int main()
{
	while (~scanf("%d%d%d", &n, &a, &b))
	{
		me(A); idx = 1;
		solve(0, (1 << n), 0, (1 << n), a - 1, b - 1);
		A[a - 1][b - 1] = 0;
		for (int i = 0; i<(1 << n); i++)
		{
			for (int j = 0; j<(1 << n); j++)
			{
				printf("%2d%c", A[i][j], " \n"[j == (1 << n) - 1]);
			}
			puts("");
		}

	}
}