poj 1185 较难的状压dp

原创于 2020-06-11 17:55:30 发布 · 248 阅读

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本文详细介绍了C++代码的优化技巧，包括宏定义、数据类型选择、算法设计等，通过具体实例展示了如何创建高效且易于维护的代码模板。

在这里插入图片描述

于6月12日更新了代码，现在代码更加容易修改，更适合作为模板。

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;
#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;

const int p[] = { 1,3,9,27,81,243,729,2187,6561,19683,59049 };

int n, m, f[111][59049 + 5], v[11];
char mp[111][11];

void mark(int last, int x) {
	for (int i = 1; i <= m; i++, last /= 3) {
		if (last % 3 == 2)v[i] = 1;
		else if (last % 3 == 1) v[i] = 0;
		else v[i] = 2;
		if (mp[x][i] == 'H'&&v[i] == 2)v[i] = 0;
	}
}

void dfs(int x, int i, int last, int now, int cnt) {
	if (i > m) {
		f[x + 1][now] = max(f[x + 1][now], f[x][last] + cnt);
		return;
	}

	if (v[i] == 2) {
		int v1 = v[i + 1], v2 = v[i + 2];
		if (v[i + 1] == 2)v[i + 1] = 0;
		if (v[i + 2] == 2)v[i + 2] = 0;
		dfs(x, i + 3, last,
			now + v[i] * p[i - 1] + v[i + 1] * p[i] + v[i + 2] * p[i + 1], cnt + 1);
		//为什么是p[i-1]而不是m-i呢？因为第i位对应的是pow(3,i-1);
		v[i + 1] = v1, v[i + 2] = v2;
		dfs(x, i + 1, last, now, cnt);
	}
	else if (v[i] == 1)
		dfs(x, i + 1, last, now + p[i - 1], cnt);
	else
		dfs(x, i + 1, last, now, cnt);
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		scanf("%s", mp[i] + 1);
	memset(f, -INF, sizeof(f));
	f[0][0] = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < 59049; j++) {
			if (f[i][j] < 0)continue;
			mark(j, i + 1);
			dfs(i, 1, j, 0, 0);
		}
	}
	int ans = 0;
	for (int i = 0; i < p[m]; i++)
		ans = max(ans, f[n][i]);
	printf("%d\n", ans);

	return 0;
}