[題目鏈接]

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

[算法]

概率DP

時間復雜度 ： O(N)

[代碼]

#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\n",dp(1,0));
        
        return 0;
    
}

[Codeforces 839C] Journey