HDU 4035 maze probability DP + tree DP

Practice: in fact, there is a root tree in the question. At first, I thought there may be multiple connected components.

The data volume is so big, and I read the blog of the great god, and slowly hurt me...

If the current node is a leaf node,

Then DP [x] = 1-E [x]-K [x] + k [x] * DP [1] + sigma (P [I] * DP [fahter [I] );

If not

DP [x] = 1-E [x]-K [x] + k [x] * DP [1] + sigma (P [I] * father [I]) + sigma (P [I] * son [I]);

Because the leaf node does not have son, and in a certain node, his son's faher is himself (nonsense,), then, in the recursion process from the leaf to the root

So DP [x] = (a [x] * DP [Father] + B [x] * DP [1] + C [x]) /(1-sigma (son (A [x]);

 

When I was doing this, # define could make me write an error ,,,

#include<cstdio>#include<cstring>#include<cmath>#define no_no 1e50#define zero(x) fabs(x)<1e-9const int LMT=100003;struct line{    int u,v;    int next;}le[LMT<<1];int next[LMT],all,du[LMT];double x[LMT],y[LMT],z[LMT],e[LMT],k[LMT];void init(void){    memset(next,-1,sizeof(next));    memset(x,0,sizeof(x));    memset(y,0,sizeof(y));    memset(z,0,sizeof(z));    memset(du,0,sizeof(du));    all=0;}void insert(int u,int v){    le[all].u=u;    le[all].v=v;    le[all].next=next[u];    next[u]=all++;}bool dfs(int u,int pre){    int v,xx;    double temy=0;    z[u]+=1-e[u]-k[u];    x[u]+=k[u];    for(xx=next[u];xx!=-1;xx=le[xx].next)        if(le[xx].v!=pre)        {            v=le[xx].v;            if(!dfs(v,u)&&1-k[u]-e[u]>0)return 0;            x[u]+=x[v]*(1-k[u]-e[u])/du[u];            temy+=y[v]*(1-k[u]-e[u])/du[u];            z[u]+=z[v]*(1-k[u]-e[u])/du[u];        }        if(pre!=-1)y[u]+=(1-k[u]-e[u])/du[u];        if(zero((temy-1.0)))return 0;            x[u]/=(1-temy);            y[u]/=(1-temy);            z[u]/=(1-temy);        return 1;}int main(void){    int T,I,i,u,v,n;    scanf("%d",&T);    for(I=1;I<=T;I++)    {        init();        scanf("%d",&n);        for(i=1;i<n;i++)        {            scanf("%d%d",&u,&v);            insert(u,v);            insert(v,u);            du[u]++;du[v]++;        }        for(i=1;i<=n;i++)        {            scanf("%lf%lf",&k[i],&e[i]);            e[i]/=100;            k[i]/=100;        }        printf("Case %d: ",I);        if(!dfs(1,-1)||zero((1.0-x[1])))            printf("impossible");        else            printf("%.6lf",z[1]/(1-x[1]));        printf("\n");    }    return 0;}