http://acm.hdu.edu.cn/showproblem.php?pid=4035

获得：

1、首先推断是不是树。事实上，所有的感觉身影，既看边数==算-1是不成立

2、有时候，我告诉孩子来区分树仍然是必要的，就是，只是是在dfs的时候，传參数的时候多加个表示父节点的參数而已

3、一定注意，概率DP对精度真的要求非常高 開始的时候写1e-8,WA了好几发，改了1e-10  AC

4、注意分母为0的可能的时候加上推断

讲的非常具体的题解：http://blog.csdn.net/morgan_xww/article/details/6776947

直接按公式写的代码就是：

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <iostream>

#include <iomanip>

#include <cmath>

#include <map>

#include <set>

#include <queue>

using namespace std;

#define ls(rt) rt*2

#define rs(rt) rt*2+1

#define ll long long

#define ull unsigned long long

#define rep(i,s,e) for(int i=s;i<e;i++)

#define repe(i,s,e) for(int i=s;i<=e;i++)

#define CL(a,b) memset(a,b,sizeof(a))

#define IN(s) freopen(s,"r",stdin)

#define OUT(s) freopen(s,"w",stdout)

const ll ll_INF = ((ull)(-1))>>1;

const double EPS = 1e-10;

const int INF = 100000000;

const int MAXN = 10000+100;

vector<int>g[MAXN];

double k[MAXN],e[MAXN];

double a[MAXN],b[MAXN],c[MAXN];

int n;

bool sea(int i, int fa)

{

    if(g[i].size() == 1 && fa!=-1)//叶子节点

    {

        a[i]=k[i];

        c[i]=b[i]=1.0-k[i]-e[i];

        return true;

    }

    //非叶子节点,此时该非叶子节点的子孙都已经遍历过了

    double aa=0.0,bb=0.0,cc=0.0;

    for(int j=0;j<g[i].size();j++)

    {

        if( g[i][j] == fa)continue;

        if(!sea(g[i][j],i))return 0;

        aa+=a[g[i][j]];

        bb+=b[g[i][j]];

        cc+=c[g[i][j]];

    }

    int m=g[i].size();

    a[i]=(k[i]+(1-k[i]-e[i])/m*aa)/(1-(1.0-k[i]-e[i])/m*bb);

    b[i]=(1.0-k[i]-e[i])/m/(1.0-(1.0-k[i]-e[i])/m*bb);

    c[i]=( (1.0-k[i]-e[i])+(1.0-k[i]-e[i])/m*cc )/(1.0 -(1.0-k[i]-e[i])/m*bb);

    return true;

}

int main()

{

    int ncase,u,v,ic=0;

    scanf("%d",&ncase);

    while(ncase--)

    {

        scanf("%d",&n);

        for(int i=1;i<=n;i++)

            g[i].clear();

        for(int i=1;i<n;i++)

        {

            scanf("%d%d",&u,&v);

            g[u].push_back(v);

            g[v].push_back(u);

        }

        for(int i=1;i<=n;i++)

        {

            scanf("%lf%lf",&k[i],&e[i]);

            k[i]/=100.0;

            e[i]/=100.0;

        }

        printf("Case %d: ",++ic);

        if(sea(1,-1) && fabs(1.0-a[1])>EPS)

            printf("%.6lf\n",c[1]/(1.0-a[1]));

        else

            printf("impossible\n");

    }

    return 0;

}

当然更好的写法还是题解上的

#include <cstdio>

#include <iostream>

#include <vector>

#include <cmath>  

using namespace std;  

const int MAXN = 10000 + 5;  

double e[MAXN], k[MAXN];

double A[MAXN], B[MAXN], C[MAXN];  

vector<int> v[MAXN];  

bool search(int i, int fa)

{

    if ( v[i].size() == 1 && fa != -1 )

    {

        A[i] = k[i];

        B[i] = 1 - k[i] - e[i];

        C[i] = 1 - k[i] - e[i];

        return true;

    }  

    A[i] = k[i];

    B[i] = (1 - k[i] - e[i]) / v[i].size();

    C[i] = 1 - k[i] - e[i];

    double tmp = 0;  

    for (int j = 0; j < (int)v[i].size(); j++)

    {

        if ( v[i][j] == fa ) continue;

        if ( !search(v[i][j], i) ) return false;

        A[i] += A[v[i][j]] * B[i];

        C[i] += C[v[i][j]] * B[i];

        tmp  += B[v[i][j]] * B[i];

    }

    if ( fabs(tmp - 1) < 1e-10 ) return false;

    A[i] /= 1 - tmp;

    B[i] /= 1 - tmp;

    C[i] /= 1 - tmp;

    return true;

}  

int main()

{

    int nc, n, s, t;  

    cin >> nc;

    for (int ca = 1; ca <= nc; ca++)

    {

        cin >> n;

        for (int i = 1; i <= n; i++)

            v[i].clear();  

        for (int i = 1; i < n; i++)

        {

            cin >> s >> t;

            v[s].push_back(t);

            v[t].push_back(s);

        }

        for (int i = 1; i <= n; i++)

        {

            cin >> k[i] >> e[i];

            k[i] /= 100.0;

            e[i] /= 100.0;

        }  

        cout << "Case " << ca << ": ";

        if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 )

            cout << C[1]/(1 - A[1]) << endl;

        else

            cout << "impossible" << endl;

    }

    return 0;

}