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  • [Codeforces 839C] Journey

    [题目链接]

             http://codeforces.com/contest/839/problem/C

    [算法]

           概率DP

           时间复杂度 : O(N)

    [代码]

            

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 100010
    
    int tot , n;
    int head[MAXN];
    bool visited[MAXN];
    double f[MAXN];
    
    struct edge
    {
            int to , nxt;
    } e[MAXN << 1];
    
    template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
    template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
    template <typename T> inline void read(T &x)
    {
        T f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
    }
    inline void addedge(int u,int v)
    {
            tot++;
            e[tot] = (edge){v,head[u]};
            head[u] = tot;
    }
    inline double dp(int u,int fa)
    {
            int cnt = 0;
            if (visited[u]) return f[u];
            for (int i = head[u]; i; i = e[i].nxt)
            {
                    int v = e[i].to;
                    if (v == fa) continue;
                    cnt++;
            }
            if (cnt == 0) 
            {
                    visited[u] = true;
                    return f[u] = 0;
            }
            f[u] = 1;
            for (int i = head[u]; i; i = e[i].nxt)
            {
                    int v = e[i].to;
                    if (v == fa) continue;
                    f[u] += 1.0 / cnt * dp(v,u);
            }
            visited[u] = true;
            return f[u];
    }
    
    int main()
    {
            
            read(n);
            for (int i = 1; i < n; i++)
            {
                    int u , v;
                    read(u); read(v);
                    addedge(u,v);
                    addedge(v,u);
            }
            memset(visited,false,sizeof(visited));
            printf("%.10lf
    ",dp(1,0));
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9746822.html
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