[HDU4035] Maze（概率DP）

HDU4035

题解

\(1.\)设对每个结点转化为：\(E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;\) $ ??$方便dfs中转移

\(2.\)推式子要考虑清楚叶节点的情况,要记得乘上概率\(\frac{1}{m}\)

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
    register LL x=0,f=1;register char c=getchar();
    while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
    while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
    return f*x;
}

const int MAXN=1e4+5;
const double eps=1e-9;

double A[MAXN],B[MAXN],C[MAXN],e[MAXN],k[MAXN];
int n,T;
vector <int> G[MAXN];

inline bool dfs(int u,int pre){
    int m=G[u].size();
    A[u]=k[u];
    B[u]=(1-k[u]-e[u])/m;
    C[u]=1-k[u]-e[u];
    double tmp=0;
    for(int i=0;i<m;i++){
        int v=G[u][i];
        if(v==pre) continue;
        if(!dfs(v,u)) return false;
        A[u]+=(1-k[u]-e[u])/m*A[v];
        C[u]+=(1-k[u]-e[u])/m*C[v];
        tmp+=(1-k[u]-e[u])/m*B[v];
    }
    if(fabs(tmp-1)<eps) return false;//如果$Ei$的系数趋向于0则无解
    A[u]/=(1-tmp);//很巧妙地对于叶子节点答案也是正确的
    B[u]/=(1-tmp);
    C[u]/=(1-tmp);
    return true;
}

int main(){
    T=read();
    for(int Case=1;Case<=T;Case++){
        n=read();
        for(int i=1;i<=n;i++) G[i].clear();
        for(int i=1;i<=n-1;i++){
            int x=read(),y=read();
            G[x].push_back(y);
            G[y].push_back(x);
        }
        for(int i=1;i<=n;i++){
            k[i]=(double)read()/100.0;
            e[i]=(double)read()/100.0;
        }
        if((!dfs(1,0))||(fabs(1-A[1])<eps)) printf("Case %d: impossible\n",Case);
        else printf("Case %d: %.6lf\n",Case,C[1]/(1-A[1]));
    }
}

[HDU4035] Maze（概率DP）

原文：https://www.cnblogs.com/lizehon/p/10485665.html