題目連結 題意：P處放大炮H不可放，黑色區域為攻擊範圍，求大炮不能相互攻擊的情況下，安裝最多的大炮數. 分析：三維狀壓DP 一開始亂寫了一通，真不知道寫的是什麼，竟然過了樣例，欺騙了自己？？？ 看了題解要提前把狀態預處理一下，其實也就60個狀態滿足，然後三維狀壓，突然靈感來了。 仔細一想，二維就夠了呀，上去就是寫一個二維狀壓？？？不過樣例？？？ $dp[i][j]= max(dp[i][j], dp[i - 1][u] + num[vis[j]]);$

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
#define met(s) memset(s, 0, sizeof(s))
#define rep(i, a, b) for(int i = a; i <= b; ++i)
template <class T> inline void scan_d(T &ret) {
char c; ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0'), c = getchar();}}
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int mod = 1e8;
const int MAXN = 110;
int dp[110][65][65], num[1 << 11];
int a[MAXN], vis[65];
int p = 0, n, m; char ch;

bool judge(int x) {
	if(x & (x << 1)) return false;
	if(x & (x << 2)) return false;
	return true;
}

inline void init() {
	for(int i = 0; i < (1 << m); ++i) {
		if(!judge(i)) continue;
		vis[p++] = i;
		int x = i;
		while(x) {
			if(x & 1) num[i]++;
			x >>= 1;
		}
	//	printf("%d %d\n",i, num[i]);
	}
}

int main() {
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	scanf("%d %d", &n, &m); init();
	for(int i = 1; i <= n; ++i) {
		getchar(); int ans = 0;
		for(int j = 1; j <= m; ++j) {
			scanf("%c", &ch);
			if(ch == 'P') ans = (ans << 1) + 1;
			else ans = (ans << 1);
		}
		a[i] = ans;
	}
	for(int i = 0; i < p; ++i) {
		if((a[1] & vis[i]) != vis[i]) continue;
		dp[1][i][i] = num[vis[i]];
	}
	for(int i = 0; i < p; ++i) {
		if((a[2] & vis[i]) != vis[i]) continue;
		for(int j = 0; j < p; ++j) {
			if(((a[1] & vis[j]) != vis[j]) || (vis[i] & vis[j])) continue;
			dp[2][i][j] = max(dp[2][i][j], dp[1][j][j] + num[vis[i]]);
		}
	}
	for(int i = 3; i <= n; ++i) {
		for(int j = 0; j < p; ++j) {
			if((a[i] & vis[j]) != vis[j]) continue;
			for(int u = 0; u < p; ++u) {
				if((vis[j] & vis[u]) || ((a[i - 1] & vis[u]) != vis[u])) continue;
				for(int v = 0; v < p; ++v) {
					if((vis[v] & vis[u]) || (vis[v] & vis[j]) || ((a[i - 2] & vis[v]) != vis[v])) continue;
					dp[i][j][u] = max(dp[i][j][u], dp[i - 1][u][v] + num[vis[j]]);
				}
			}
		}
	}
	int sum = 0;
	for(int i = 0; i < p; ++i) {
		for(int j = 0; j < p; ++j) {
			sum = max(sum, dp[n][i][j]);
		}
	}
	printf("%d\n", sum);
	return 0;
}