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]));elseprintf("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;
}