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  • BZOJ4182 Shopping(点分治+树形dp)

      点分治,每次考虑包含根的连通块,做树形多重背包即可,dfs序优化。注意题面给的di范围是假的,坑了我0.5h,心态炸了。

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define ll long long
    #define N 510
    #define M 4010
    #define inf 1000000000
    char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
    int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
    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;
    }
    int n,m,w[N],c[N],d[N],p[N],t,cnt,ans;
    int size[N],dfn[N],id[N],f[N][M];
    bool flag[N];
    struct data{int to,nxt;
    }edge[N<<1];
    void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;} 
    void getsize(int k,int from)
    {
        size[k]=1;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from&&!flag[edge[i].to]) 
        {
            getsize(edge[i].to,k);
            size[k]+=size[edge[i].to];
        }
    }
    int findroot(int k,int from,int s)
    {
        int mx=0;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from&&!flag[edge[i].to]&&size[edge[i].to]>size[mx]) mx=edge[i].to;
        if ((size[mx]<<1)>s) return findroot(mx,k,s);
        else return k;
    }
    void dfs(int k,int from)
    {
        dfn[k]=++cnt;size[k]=1;id[cnt]=k;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from&&!flag[edge[i].to])
        {
            dfs(edge[i].to,k);
            size[k]+=size[edge[i].to];
        }
    }
    void solve(int k)
    {
        getsize(k,k);
        flag[k=findroot(k,k,size[k])]=1;
        cnt=0;dfs(k,k);
        memset(f[size[k]+1],0,sizeof(f[size[k]+1]));
        //for (int i=1;i<=n;i++) cout<<size[i]<<' ';cout<<endl;
        //for (int i=1;i<=n;i++) cout<<dfn[i]<<' ';cout<<endl;
        //for (int i=1;i<=n;i++) cout<<id[i]<<' ';cout<<endl;
        //cout<<endl;
        for (int i=size[k];i;i--)
        {
            int x=id[i];
            for (int j=0;j<=m;j++)
            if (j>=c[x]) f[i][j]=f[i+1][j-c[x]]+w[x];
            else f[i][j]=-inf;
            int y=d[x];
            for (int u=0;u<=13;u++)
            if (y)
            {
                int z=min(y,1<<u);
                for (int j=m;j>=c[x]*z;j--)
                f[i][j]=max(f[i][j],f[i][j-c[x]*z]+w[x]*z);
                y-=z;
            }
            for (int j=0;j<=m;j++)
            f[i][j]=max(f[i][j],f[i+size[x]][j]);
        }
        ans=max(ans,f[1][m]);
        for (int i=p[k];i;i=edge[i].nxt)
        if (!flag[edge[i].to]) solve(edge[i].to);
    }
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("bzoj4182.in","r",stdin);
        freopen("bzoj4182.out","w",stdout);
        const char LL[]="%I64d
    ";
    #else
        const char LL[]="%lld
    ";
    #endif
        int T=read();
        while (T--)
        {
            n=read(),m=read();memset(p,0,sizeof(p));t=ans=0;
            for (int i=1;i<=n;i++) w[i]=read();
            for (int i=1;i<=n;i++) c[i]=read();
            for (int i=1;i<=n;i++) d[i]=read()-1;
            for (int i=1;i<n;i++)
            {
                int x=read(),y=read();
                addedge(x,y),addedge(y,x);
            }
            memset(flag,0,sizeof(flag));solve(1);
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/10287025.html
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