HDU 5917 Instability ramsey定理

http://acm.hdu.edu.cn/showproblem.php?pid=5917

即世界上任意6个人中，总有3个人相互认识，或互相皆不认识。

所以子集 >= 6的一定是合法的。

然后总的子集数目是2^n，减去不合法的，暴力枚举即可。

选了1个肯定不合法，2个也是，3个的话C(n, 3)枚举判断，C(n, 4)， C(n, 5)

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 50 + 20;
int e[maxn][maxn];
const LL MOD = 1e9 + 7;
void add(LL &x, LL y) {
    x = (x + MOD + y) % MOD;
}
bool ok(int i, int j, int k) {
    return ((e[i][j] && e[j][k] && e[k][i]) || (!e[i][j] && !e[j][k] && !e[k][i]));
}
LL po[maxn];
int f;
void work() {
    memset(e, false, sizeof e);
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        e[u][v] = e[v][u] = 1;
    }
    LL ans = po[n];
    add(ans, -(n + 1));
    add(ans, -(n * (n - 1) / 2));
    LL t = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            for (int k = j + 1; k <= n; ++k) {
                if (!ok(i, j, k)) t++;
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            for (int k = j + 1; k <= n; ++k) {
                for (int h = k + 1; h <= n; ++h) {
                    if (!ok(i, j, k) && !ok(i, j, h) && !ok(i, k, h) && !ok(j, k, h)) t++;
                }
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = i + 1; j <= n; ++j) {
            for (int k = j + 1; k <= n; ++k) {
                for (int h = k + 1; h <= n; ++h) {
                    for (int ha = h + 1; ha <= n; ++ha) {
                        if (!ok(i, j, k) && !ok(i, j, h) && !ok(i, k, h) && !ok(j, k, h) && !ok(i, j, ha) &&
                          !ok(i, k, ha) && !ok(i, h, ha) && !ok(j, k, ha) && !ok(j, h, ha) &&
                          !ok(k, h, ha)) t++;
                    }
                }
            }
        }
    }
    t %= MOD;
    add(ans, -t);
    printf("Case #%d: %lld
", ++f, ans);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    po[0] = 1;
    for (int i = 1; i <= maxn - 2; ++i) {
        po[i] = 2 * po[i - 1] % MOD;
    }
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}