uva 1390 - Interconnect(期望+哈希+记忆化)

题目连接：uva 1390 - Interconnect

题目大意：给出n表示有n个点，m表示有m条边，如今任选两点建立一条边。直到整个图联通，问说还需建立边数的期望，建过边的两点仍能够建边。

解题思路：哈希的方法非常是巧妙。将各个联通分量中节点的个数c[i]转换成一个30进制的数（由于节点个数最多为30）。由于结果非常大。所以对1e5+7取模。获得的哈希值作为插入和搜索的起点。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 30;
const int mod = 1e5+7;

struct state {
    int c[maxn], flag;
    double val;

    void clear () { memset(c, 0, sizeof(c)); }
    int hash() {
        int x = 0;
        for (int i = 0; i < maxn; i++)
            x = (x * 30 + c[i]) % mod;
        return x;
    }
    bool operator == (const state& u) {
        for (int i = 0; i < maxn; i++)
            if (c[i] != u.c[i])
                return false;
        return true;
    }

    bool operator != (const state& u) {
        return !(*this == u);
    }

}start, ha[mod+7];

int n, m, f[maxn+5], s[maxn+5];
double dive;

int getfar (int x) {
    return f[x] == x ? x : f[x] = getfar(f[x]);
}

void link (int x, int y) {
    int p = getfar(x);
    int q = getfar(y);

    if (p == q)
        return;

    f[q] = p;
    s[p] += s[q];
}

void inserthash (state u) {
    int x = u.hash();
    while (ha[x].flag) {
        if (++x == mod)
            x = 0;
    }
    ha[x] = u;
    ha[x].flag = 1;
}

double gethash (state u) {
    int x = u.hash();
    while (ha[x].flag && ha[x] != u) {
        if (++x == mod)
            x = 0;
    }
    return ha[x] == u ? ha[x].val : -1;
}

void init () {
    dive = n * (n - 1) / 2.0;
    start.clear();
    for (int i = 0; i <= n; i++) {
        s[i] = 1;
        f[i] = i;
    }

    for (int i = 0; i < mod; i++)
        ha[i].flag = 0;

    int a, b;
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &a, &b);
        link(a, b);
    }

    for (int i = 1; i <= n; i++) {
        if (f[i] == i)
            start.c[i-1] = s[i];
    }
}

double solve (state u) {
    sort(u.c, u.c+maxn);
    if (u.hash() == n)
        return 0;

    double x = gethash(u);

    if (x != -1.0)
        return x;

    double ans = 0, tmp = 0;

    for (int i = 0; i < maxn; i++)
        tmp += u.c[i] * (u.c[i] - 1) / 2.0;

    for (int i = 0; i < maxn; i++) {

        if (u.c[i] == 0)
            continue;

        for (int j = i+1; j < maxn; j++) {

            if (u.c[j] == 0)
                continue;

            state v = u;
            v.c[i] += v.c[j];
            v.c[j] = 0;
            ans += u.c[i] * u.c[j] * solve(v);
        }
    }

    ans /= dive;
    ans++;
    ans /= (1 - tmp / dive);
    u.val = ans;
    inserthash(u);
    return ans;
}

int main () {
    while (scanf("%d%d", &n, &m) == 2) {
        init();
        printf("%.10lf\n", solve(start));
    }
    return 0;
}