解題區/解題區 - UVa

UVa 1390 - Interconnect

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

N 個城鎮之間一開始有一些邊相連，接著每次任挑兩個城鎮拉一條邊相連，請問期望值需要幾次才能將 N 個城鎮彼此相連成為一個連通元件。

原本彼此連通的城鎮，也有可能重複拉邊。

Sample Input

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2 1 
1 2 
4 2 
1 2 
3 4

Sample Output

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2
0.0 
1.5

Solution

事實上，將狀態壓縮成每一個連通元件的節點個數，對於連通元件之間做拉邊即可。

對於狀態 u, v，期望值 E[u] = 1 + E[v] * p + E[u] * q，

E[v] * p 表示：將兩個不同連通元件之間拉邊，則會將兩個連通元件融合，機率 p。
E[u] * q 表示：在各自的連通元件內布拉邊，不影響狀態結果。

左移一下得到 E[u] = (1 + E[v] * p) / (1 - q);

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#include <stdio.h> 
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
int parent[32], weight[32];
int findp(int x) {
    return parent[x] == x ? x : findp(parent[x]);
}
void joint(int x, int y) {
    x = findp(x), y = findp(y);
    if (x == y)	return;
    if (weight[x] > weight[y])
        parent[y] = x, weight[x] += weight[y];
    else
        parent[x] = y, weight[y] += weight[x];
}
#define hash_mod 1000003
struct state {
    vector<int> c;
    unsigned int hash() {
        unsigned int a = 63689, b = 378551;
        unsigned int value = 0;
        for (int i = 0; i < c.size(); i++) {
            value = value * a + c[i];
            a *= b;
        }
        return value % hash_mod;
    }
    bool operator<(const state &a) const {
        if (c.size() != a.c.size())
            return c.size() < a.c.size();
        for (int i = 0; i < c.size(); i++)
            if (c[i] != a.c[i])
                return c[i] < a.c[i];
        return false;
    }
};
map<state, double> hash[hash_mod];
double dfs(state &u) {
    sort(u.c.begin(), u.c.end());
    if (u.c.size() == 1)	return 0;
    
    int h = u.hash();
    if (hash[h].find(u) != hash[h].end())
        return hash[h][u];
    double &ret = hash[h][u];
    double self = 0, total = 0;
    ret = 0;
    for (int i = 0; i < u.c.size(); i++) {
        self += u.c[i] * (u.c[i] - 1) / 2.0;
        total += u.c[i];
    }
    total = total * (total - 1) / 2.0;
    for (int i = 0; i < u.c.size(); i++) {
        for (int j = i+1; j < u.c.size(); j++) {
            state v = u;
            v.c[i] += v.c[j];
            v.c.erase(v.c.begin() + j);
            ret += u.c[i] * u.c[j] * dfs(v);
        }
    }
    // E[u] = 1 + E[v] * p + E[u] * q
    ret = (ret / total) + 1;
    ret = ret / (1 - self / total);
    return ret;
}
int main() {
    int n, m, x, y;
    while (scanf("%d %d", &n, &m) == 2) {
        for (int i = 1; i <= n; i++)
            parent[i] = i, weight[i] = 1;
        for (int i = 0; i < m; i++) {
            scanf("%d %d", &x, &y);
            joint(x, y);
        }
        
        state st;
        for (int i = 1; i <= n; i++) {
            if (parent[i] == i) // root
                st.c.push_back(weight[i]);
        }
        printf("%lf\n", dfs(st));
    }
    return 0;
}