Bzoj 2878: [noi2012] lost amusement park

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Question:

A weighted undirected graph with n points n-1 or n edges is provided. Start from each point and never go through another point. What is the mathematical expectation of the path length?

N <= 100000. In the figure, there is at most one ring, and the length of the Ring cannot exceed 20.

Question:

First, consider a tree:

First, take a node as the root node (for example, point 1) and create a tree chart.

The expected length of the Child tree where I goes is f [I], and the child nodes of I are S1, S2 ...... SK.

If d [I] = Σ (F [SJ] + edge (I, SJ) is set, F [I] = d [I]/K.

If P [I] is set to I as the expected length of the start point, P [1] = f [1] and Du [I] is the degree of node I.

Run the DFS again. If Y is a subnode of X, p [x] has been calculated.

The expected contribution of node y to X is other, other = (P [x] * Du [x]-edge (x, y)-f [x]). /(Du [x]-1) + edge (x, y)

P [y] = (Other + d [y])/du [y]

In this way, the P values of all vertices can be calculated through two DFS operations.

However, this question may be a base ring tree.

First, you can find this ring through DFS. If you remove this ring, it is the forest composed of some sub-trees.

First, use the aforementioned tree-like motion gauge method to process these sub-trees and find D and F.

Then we noticed that there were few points on the ring, so we calculated each point on the ring as the starting point.

For point X on the ring, set calc_pre (X) to the expected length from X to the left side of the ring, calc_nxt (X) the expected length obtained from the right side of the X-ring.

If SX is X.

Then P [x] = (Σ (F [SX] + edge (x, SX) + calc_pre (x) + calc_nxt (x)/du (X ).

Among them, calc_pre and calc_nxt can be calculated based on the complexity of O (number of points on the ring.

So we can calculate the p of all points in DFS again.

Ans = (Σ p [I])/n

/**************************************************************    Problem: 2878    User: zrts    Language: C++    Result: Accepted    Time:588 ms    Memory:10748 kb****************************************************************/ #include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>//by zrt//problem:using namespace std;typedef long long LL;const int inf(0x3f3f3f3f);const double eps(1e-9);bool oncircle[100005];int du[100005];int fa[100005];int H[100005],X[200005],P[200005],tot;double E[200005];inline void add(int x,int y,int z){//add an edge    P[++tot]=y;X[tot]=H[x];H[x]=tot;E[tot]=z;}int n,m;double F[100005];//for son Edouble D[100005];// F * |son|double p[100005];// ansint num[100005];// |son|int pre[100005];double wpre[100005];bool type;void treedp1(int x){//Dp the tree first    for(int i=H[x];i;i=X[i]){        if(P[i]==fa[x]||oncircle[P[i]]) continue;        fa[P[i]]=x;        treedp1(P[i]);        D[x]+=F[P[i]]+E[i];num[x]++;    }    if(num[x])    F[x]=D[x]/num[x];    else F[x]=0;}void treedp2(int x,double edge){//Dp the tree second | calc the P    double other;    if(!fa[fa[x]]) {        if(!type){            if(num[fa[x]]>1) other=(p[fa[x]]*(num[fa[x]]+0)-edge-F[x])/(num[fa[x]]-1)+edge;            else other=edge;        }else{            other=(p[fa[x]]*(num[fa[x]]+2)-edge-F[x])/(num[fa[x]]+1)+edge;        }    }    else other=(p[fa[x]]*(num[fa[x]]+1)-edge-F[x])/num[fa[x]]+edge;         p[x]=(other+D[x])/(num[x]+1);    for(int i=H[x];i;i=X[i]){        if(P[i]==fa[x]||oncircle[P[i]]) continue;        treedp2(P[i],E[i]);    }}vector<int> circle;bool ok;int nxt[100005];double wnxt[100005];void findcircle(int x,int fa){    for(int i=H[x];i;i=X[i]){        if(P[i]==fa) continue;        if(pre[P[i]]){            ok=1;            int k=x;            pre[P[i]]=x;            wpre[P[i]]=E[i];            do{                oncircle[k]=1;                circle.push_back(k);                k=pre[k];            }while(k!=P[i]);            oncircle[k]=1;            circle.push_back(k);        }else{            pre[P[i]]=x;wpre[P[i]]=E[i];            findcircle(P[i],x);            if(ok) return;        }    }}double calc_nxt(int now,int end){    double ret=wnxt[now];    now=nxt[now];    int extra=0;    if(nxt[now]!=end) ret+=calc_nxt(now,end)/(du[now]-1);    else extra=-1;    for(int i=H[now];i;i=X[i]){        if(oncircle[P[i]]) continue;        ret+=(E[i]+F[P[i]])/(du[now]-1+extra);    }    return ret;}double calc_pre(int now,int end){    double ret=wpre[now];    now=pre[now];    int extra=0;    if(pre[now]!=end) ret+=calc_pre(now,end)/(du[now]-1);    else extra=-1;    for(int i=H[now];i;i=X[i]){        if(oncircle[P[i]]) continue;        ret+=(E[i]+F[P[i]])/(du[now]-1+extra);    }    return ret;}int main(){// the main function    #ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    #endif    scanf("%d%d",&n,&m);    for(int i=0,x,y,z;i<m;i++){        scanf("%d%d%d",&x,&y,&z);        du[x]++;        du[y]++;        add(x,y,z);        add(y,x,z);    }    if(m==n-1){        treedp1(1);        p[1]=F[1];        for(int i=H[1];i;i=X[i]){            treedp2(P[i],E[i]);        }        double ans=0;        for(int i=1;i<=n;i++){             ans+=p[i];        }        printf("%.5f\n",ans/n);    }else{        type=1;        pre[1]=-1;        findcircle(1,-1);        for(int now=0;now<circle.size();now++){            nxt[pre[circle[now]]]=circle[now];            wnxt[pre[circle[now]]]=wpre[circle[now]];        }        for(int now=0;now<circle.size();now++){            for(int i=H[circle[now]];i;i=X[i]){                if(oncircle[P[i]]) continue;                treedp1(P[i]);                p[circle[now]]+=(F[P[i]]+E[i])/du[circle[now]];                num[circle[now]]++;            }        }        for(int now=0;now<circle.size();now++){            p[circle[now]]+=(calc_pre(circle[now],circle[now])+calc_nxt(circle[now],circle[now]))/du[circle[now]];        }        for(int now=0;now<circle.size();now++){            for(int i=H[circle[now]];i;i=X[i]){                if(oncircle[P[i]]) continue;                fa[P[i]]=circle[now];                treedp2(P[i],E[i]);            }        }        double ans=0;        for(int i=1;i<=n;i++){            ans+=p[i];        }        printf("%.5f\n",ans/n);    }    return 0;}

Bzoj 2878: [noi2012] lost amusement park