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P2704 [NOI2001]炮兵阵地 (状压DP)

题目:

P2704 [NOI2001]炮兵阵地

解析:

和互不侵犯一样
就是多了一格
\(f[i][j][k]\)表示第i行,上一行状态为\(j\),上上行状态为\(k\)的最多的可以放的炮兵
发现\(100\times 1024\times 1024\)开不下
还是通过简单的搜索发现就算\(m==10\)时合法的状态只有\(60\)
\(100\times 60\times 60\)就没问题了
然后就和互不侵犯一样,枚举状态就可以了
状态转移
\(f[i][j][k] = max\{f[i][j][k], f[i-1][k][l]+sum[state[i]]\}\)
\(sum[i]\)表示\(i\)状态中有多少个\(1\)

代码:

#include <bits/stdc++.h>
using namespace std;
const int N = 110;

int n, m, num, ans = -0x3f3f3f3f;
int state[N], sum[1050], line[N], f[N][N][N];

char s[N];

int qpow(int a, int b) {
	int ans = 1;
	while (b) {
		if (b & 1) ans = ans * a;
		a *= a, b >>= 1;
	}
	return ans;
}

int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		cin >> s;
		for (int j = m - 1; j >= 0; --j)
			if (s[j] == 'P') line[i] += qpow(2, m - j - 1);
	}
	for (int i = 0; i < (1 << m); ++i) {
		if ((i & (i << 1)) || (i & (i >> 1)) ||
		    (i & (i << 2)) || (i & (i >> 2))) continue;
		state[++num] = i;
		sum[i] = __builtin_popcount(i);
	}
	for (int i = 1; i <= num; ++i)
		if ((state[i] | line[1]) == line[1])
			f[1][i][1] = sum[state[i]];
	for (int i = 2; i <= n; ++i)
		for (int j = 1; j <= num; ++j)
			if ((state[j] | line[i]) == line[i])
				for (int k = 1; k <= num; ++k)
					if ((state[k] | line[i - 1]) == line[i - 1] && !(state[j] & state[k]))
						for (int l = 1; l <= num; ++l)
							if (!(state[l] & state[k]) &&
							    !(state[l] & state[j]) &&
							    (state[l] | line[i - 2]) == line[i - 2])
								f[i][j][k] = max(f[i][j][k], f[i - 1][k][l] + sum[state[j]]);
	for (int i = 1; i <= num; ++i)
		for (int j = 1; j <= num; ++j)
			ans = max(ans, f[n][i][j]);
	cout << ans;
}
posted @ 2019-10-11 16:49  Chrety  阅读(177)  评论(5编辑  收藏