Ultraviolet A 1390-Interconnect (expect + hash + Memory)

Question connection: Ultraviolet A 1390-Interconnect

N indicates that there are n vertices and M indicates that there are m edges. Now, you can set up an edge at either of the two points until the entire graph is connected and you need to set up the expected number of edges, two points of edge creation can still be used.

Solution: the hash method is very clever. It converts the number of nodes in each Unicom component C [I] into a 30-digit number (because the number of nodes is up to 30 ), the result is very large, so the modulo of 1e5 + 7 is obtained. The obtained hash value is used as the starting point for insertion and search.

#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;}