【APIO2007】动物园

Description

Solution

每个人能看到的动物只有5个，所以我们可以将这个进行状压，定义f[i][j]表示从第i开始的五个动物状态压缩后（移走或不移走）为j的最优结果

那么转移方程可以得到

其中，val表示这一步转移所获得的收益，由于动物园是环形的，所以初状态等于末状态，我们枚举末状态，取出最大值即可

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read() {
    int ret = 0;
    int op = 1;
    char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') op = -1; c = getchar();}
    while (c <= '9' && c >= '0') ret = ret * 10 + c - '0', c = getchar();
    return ret * op;
}
int n, m, ans;
int f[10010][1 << 6], val[10010][1 << 6];
int main() {
    n = read(); m = read();
    for (int i = 1; i <= m; ++i) {
        int a = read(), b = read(), c = read();
        int x = 0, y = 0, now;
        for (int j = 1; j <= b; ++j) {
            now = read();
            now = (now - a + n) % n;
            x |= 1 << now;
        }
        for (int j = 1; j <= c; ++j) {
            now = read();
            now = (now - a + n) % n;
            y |= 1 << now;
        }
        for (int j = 0; j < (1 << 5); ++j)
            if ((j & x) || ((~j) & y)) val[a][j]++;
    }
    for (int i = 0; i < (1 << 5); ++i) {
        memset(f[0], 128, sizeof(f[0]));
        f[0][i] = 0;
        for (int j = 1; j <= n; ++j) {
            for (int s = 0; s < (1 << 5); ++s)
                f[j][s] = max(f[j - 1][(s & 15) << 1], f[j - 1][(s & 15) << 1 | 1]) + val[j][s];
        }
        ans = max(ans, f[n][i]);
    }
    printf("%d
", ans);
    return 0;
}