[bzoj1151][CTSC2007]动物园zoo_状压dp

动物园zoo

题目大意：https://www.lydsy.com/JudgeOnline/problem.php?id=1151

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题解：

我们发现每个点只会往右延伸$5$个，这个数非常小。

再加上每个动物只有选和不选，很容易想到把每个点后面$5$个给状压到一起。

想到这里就好办了，随便弄个数组搞一搞就好。

代码：

#include <bits/stdc++.h>

#define N 50010 

using namespace std;

int n, c, f[N][35], bu[N][35], last, ans;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
    int x = 0, f = 1;
    char c = nc();
    while (c < 48) {
        if (c == '-')
            f = -1;
        c = nc();
    }
    while (c > 47) {
        x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
    }
    return x * f;
}

int main() {
    int n = rd(), c = rd();
    for (int i = 1; i <= c; i ++ ) {
        int a = rd(), l1 = rd(), l2 = rd();
        int t1 = 0, t2 = 0;
        for (int j = 1; j <= l1; j ++ ) {
            t1 |= 1 << ((rd() - a + n) % n);
        }
        for (int j = 1; j <= l2; j ++ ) {
            t2 |= 1 << ((rd() - a + n) % n);
        }
        for (int j = 0; j < 32; j ++ ) {
            if ((j & t1) || (~j & t2)) {
                bu[a][j] ++ ;
            }
        }
    }
    last = 0;
    for (int j = 0; j < 32; j ++ ) {
        memset(f[0], 0xef, sizeof f[0]);
        f[0][j] = 0;
        for (int i = 1; i <= n; i ++ ) {
            for (int k = 0; k < 32; k ++ ) {
                f[i][k] = max(f[i - 1][(k & 15) << 1], f[i - 1][(k & 15) << 1 | 1]) + bu[i][k];
            }
        }
        ans = max(ans, f[n][j]);
    }
    cout << ans << endl ;
    return 0;
}

小结：发现前i个只跟后面5个的01情况有关，我们就直接存起来f[i][S]就好。