算法：轮廓线dp

代码：
#include<iostream>
#include<stack>
#include<string.h>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<climits>
#include<cstdio>
#include<vector>
#include<sstream>
#include<ctype.h>
#include<set>
#include<map>
#include<ctime>
#include<stdlib.h>
#include<queue>
#include<bitset>
#define ll long long
#define inf INT_MAX
using namespace std;
template<typename TTT>inline void mr(TTT& theNumberToRead)
{
	theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	if (prn)
		theNumberToRead = -theNumberToRead;
}
template<typename TTT>inline TTT mrr()
{
	TTT theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	return prn ? -theNumberToRead : theNumberToRead;
}
template<typename T>void my_write(T x)
{
	if (x)
		my_write(x / 10),
		putchar(x % 10 ^ 48);
}
template<typename T>void mw(T x, char mid)
{
	if (x)
	{
		if (x < 0)
			putchar('-'),
			x = -x;
		my_write(x);
	}
	else putchar(48);
	if (mid)
		putchar(mid);
}
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
ll dp[2][1 << 11];
int main()
{
	int n, m;
	while (mr(n), mr(m), n)
	{
		if (n < m)
			swap(n, m);
		memset(dp[0], 0, sizeof(dp));
		ll* crt = dp[0], * nex = dp[1];
		crt[0] = 1;
		for(int i=n-1; ~i; --i)
			for (int j = m - 1; ~j; --j)
			{
				for (int k = 0; k < 1 << m; ++k)
				{
					if (k & 1 << j)
						nex[k] = crt[k ^ 1 << j];
					else
					{
						nex[k] = 0;
						if (j + 1 < m && !(k & 1 << j + 1))
							nex[k] += crt[k | 1 << j + 1];
						if (i + 1 < n)
							nex[k] += crt[k | 1 << j];
					}
				}
				swap(crt, nex);
			}
		mw(crt[0], '\n');
	}
	return 0;
}

• Re:《挑战程序设计竞赛》的解法，复杂度nm2^m (30) Ioencgc 2023-07-21 12:24:16