HDU 4035 possibilities DP Chengdu online game

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

Get:

1, the first inference is not a tree. In fact, all the feeling figure, both the look at the number of sides = = count-1 is not set

2, sometimes, I tell the child to distinguish the tree is still necessary, that is, only when the DFS, when the number of references to a parent node to indicate the number of parameters

3, must pay attention to the probability of DP to the accuracy of the very high need to start writing 1e-8,wa several, changed 1e-10 AC

4, note that the denominator is 0 possible when adding inference

A very specific explanation: http://blog.csdn.net/morgan_xww/article/details/6776947

The code written directly by the formula is:

#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#de Fine 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&LT;INT&GT;G[MAXN];d ouble K[MAXN],E[MAXN];d o Uble a[maxn],b[maxn],c[maxn];int n;bool Sea (int i, int fa) {if (g[i].size () = = 1 && fa!=-1)//leaf node {a[        I]=k[i];        C[i]=b[i]=1.0-k[i]-e[i];    return true;    }//non-leaf node, at this time the descendants of the non-leaf node have traversed 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)/(n (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;}

Of course, the better writing is still a puzzle.

#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;  }

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HDU 4035 possibilities DP Chengdu online game