POJ1185 - 炮兵阵地_炮兵阵地 dp yxpandjay-优快云博客

本文深入探讨状压DP（Bitmask Dynamic Programming）算法的基础概念及应用，通过实例演示如何利用位运算技巧简化状态转移方程，有效解决具有大量状态的选择问题。从基本原理出发，逐步引导读者理解并掌握状压DP的核心思想，同时展示其在实际编程场景中的高效实现。

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入门状压DP

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define INF 1000000009
#define _ std::ios_base::sync_with_stdio(0);cin.tie(0);
#define REP(i,nn) for (int i = 0; i < int(nn); ++ i)
#define REPI(i,aa,nn) for (int i = int(a); i < int(nn); ++ i)
#define maxn  19
#define MOD ((int)1e9+7)
using namespace std;
typedef long long ll;
int cur[110], state[110], dp[110][110][110], num[110];
int count(int x) {
	int cnt = 0;
	while (x) {
		x &= (x-1);
		cnt ++;
	}
	return cnt;
}
int main() {
	int n, m, top = 0;
	while (cin >> n >> m) {
		for (int i = 0; i < n; ++ i) {
			cur[i] = 0;
			for (int j = 0; j < m; ++ j) {
				char t;
				cin >> t;
				if ( t == 'H')
					cur[i] += (1 << (m-j-1));
			}
			//cout << cur[i] << endl;
		}
		for (int i = 0; i < (1<<m); ++ i) {
			if (!(i&(i<<1)) && !(i&(i<<2))) {
				state[top++] = i;
			}
		}
		memset(dp, -1, sizeof(dp));
		for (int i = 0; i < top; ++ i) {
			num[i] = count(state[i]);
			if (!(state[i]&cur[0]))
				dp[0][0][i] = num[i];
		}
		//for (int i = 0; i < top; ++ i) cout << num[i] << " ";
		//cout << endl;
		for (int i = 1; i < n; ++ i) {
			for (int t = 0; t < top; ++ t) {
				if (state[t]&cur[i]) continue;
				for (int j = 0; j < top; ++ j) {
					if (state[j]&state[t]) continue;
					for (int k = 0; k < top; ++ k) {
						if (state[k]&state[t]) continue;
						if (dp[i-1][j][k] == -1) continue;
						//puts("ok");
						dp[i][k][t] = max(dp[i][k][t], dp[i-1][j][k] + num[t]);
					}
				}
			}
		}
		int res = 0;
		for (int i = 0; i < n; ++ i)
		for (int j = 0; j < top; ++ j)
		for (int k = 0; k < top; ++ k)
			res = max(res, dp[i][j][k]);
		cout << res << endl;
	}
}