BZOJ4446 SCOI2015小凸玩密室（树形dp）

 　　设f[i][j]为由根进入遍历完i子树，最后一个到达的点是j时的最小代价，g[i][j]为由子树内任意一点开始遍历完i子树，最后一个到达的点是j时的最小代价，因为是一棵完全二叉树，状态数量是nlogn的。转移考虑四种走法：根→左子树→右子树；根→右子树→左子树；左子树→根→右子树；右子树→根→左子树 即可。好像和大多数题解不一样，明明感觉这种做法比较显然。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 200010
#define inf 100000000000000000ll
int n,a[N],p[N],deep[N],t;
vector<int> son[N];
vector<long long> f[N],g[N];
struct data{int to,nxt,len;
}edge[N];
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
void dfs(int k)
{
    int cnt=0;
    for (int i=p[k];i;i=edge[i].nxt)
    {
        cnt++;
        deep[edge[i].to]=deep[k]+edge[i].len;
        dfs(edge[i].to);
        for (int j=0;j<son[edge[i].to].size();j++)
        son[k].push_back(son[edge[i].to][j]);
    }
    son[k].push_back(k);
    if (cnt==0) {f[k].push_back(0);g[k].push_back(0);return;}
    if (cnt==1)
    {
        long long v=inf;
        for (int i=0;i<son[k].size()-1;i++)
        f[k].push_back(f[edge[p[k]].to][i]+1ll*edge[p[k]].len*a[edge[p[k]].to]),
        g[k].push_back(f[k][i]),v=min(v,f[edge[p[k]].to][i]+1ll*(deep[son[edge[p[k]].to][i]]-deep[k])*a[k]);
        f[k].push_back(inf),g[k].push_back(v);
    }
    else
    {
        long long P=inf,Q=inf,X=inf,Y=inf;
        int x=edge[p[k]].to,y=edge[edge[p[k]].nxt].to,w=edge[p[k]].len,v=edge[edge[p[k]].nxt].len;
        for (int i=0;i<son[x].size();i++)
        P=min(P,f[x][i]+1ll*w*a[x]+1ll*(deep[son[x][i]]-deep[k])*a[y]),
        X=min(X,g[x][i]+1ll*(deep[son[x][i]]-deep[k])*a[k]);
        for (int i=0;i<son[y].size();i++)
        Q=min(Q,f[y][i]+1ll*v*a[y]+1ll*(deep[son[y][i]]-deep[k])*a[x]),
        Y=min(Y,g[y][i]+1ll*(deep[son[y][i]]-deep[k])*a[k]);
        X=min(X,P),Y=min(Y,Q);
        for (int i=0;i<son[x].size();i++)
        f[k].push_back(Q+1ll*w*a[x]+f[x][i]),g[k].push_back(Y+1ll*w*a[x]+f[x][i]);
        for (int i=0;i<son[y].size();i++)
        f[k].push_back(P+1ll*v*a[y]+f[y][i]),g[k].push_back(X+1ll*v*a[y]+f[y][i]);
        f[k].push_back(inf),g[k].push_back(inf);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4446.in","r",stdin);
    freopen("bzoj4446.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<n;i++) addedge(i+1>>1,i+1,read());
    dfs(1);
    long long ans=inf;
    for (int i=0;i<son[1].size();i++) ans=min(ans,g[1][i]);
    cout<<ans;
    return 0;
}