BZOJ4860 Beijing2017树的难题（点分治+单调队列）

　　考虑点分治。对子树按照根部颜色排序，每次处理一种颜色的子树，对同色和不同色两种情况分别做一遍即可，单调队列优化。但是注意到这里每次使用单调队列的复杂度是O(之前的子树最大深度+该子树深度)，一不小心就退化成O(n²)。于是我们按照同颜色最大深度为第一关键字、子树深度为第二关键字排序，每次处理完一种颜色再与之前的其他颜色合并，这样每次的复杂度就是其自身深度了。

#include 
#include
#include
#include
#include
#include
#include
using namespace std;
#define ll long long
#define N 200010
#define inf 2100000000
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,l,r,p[N],a[N],mxdeep[N],colordeep[N],deep[N],size[N],len[N],f[N],g[N],q[N][2],Q[N],t,ans=-inf;
bool flag[N];
struct data{int to,nxt,color;
}edge[N<<1];
struct data2
{
    int x,y;
    bool operator <(const data2&a) const
    {
        return colordeep[y]mxdeep[a.x];
    }
}v[N];
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].color=z,p[x]=t;}
void make(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])
    {
        deep[edge[i].to]=deep[k]+1;
        make(edge[i].to,k);
        size[k]+=size[edge[i].to];
    }
}
int findroot(int k,int s)
{
    int mx=0;
    for (int i=p[k];i;i=edge[i].nxt) 
    if (size[edge[i].to]size[mx]&&!flag[edge[i].to]) mx=edge[i].to;
    if ((size[mx]<<1)>s) return findroot(mx,s);
    else return k;
}
void findmx(int k,int from,int root)
{
    mxdeep[root]=max(mxdeep[root],deep[k]);
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from&&!flag[edge[i].to]) findmx(edge[i].to,k,root);
}
inline void ins(int k,int &head,int &tail,int *f){while (head<=tail&&f[Q[tail]]k;}
inline void pop(int k,int &head,int &tail){while (head<=tail&&Q[head]>k) head++;}
void bfs(int k,int last,int v,int d)
{
    int head=0,tail=1;q[1][0]=k,q[1][1]=last;
    int h=1,t=0;for (int i=min(d,r);i>=l;i--) ins(i,h,t,g);
    do
    {
        int x=q[++head][0];if (deep[x]>=l&&deep[x]<=r) ans=max(ans,len[x]);
        if (deep[x]>deep[q[head-1][0]])
        {
            pop(r-deep[x],h,t);
            if (l-deep[x]>=0&&l-deep[x]<=d) ins(l-deep[x],h,t,g);
        }
        if (h<=t) ans=max(ans,len[x]+g[Q[h]]+v);
        for (int i=p[x];i;i=edge[i].nxt)
        if (deep[edge[i].to]>deep[x]&&!flag[edge[i].to])
        {
            len[edge[i].to]=len[x];
            if (edge[i].color!=q[head][1]) len[edge[i].to]+=a[edge[i].color];
            q[++tail][0]=edge[i].to,q[tail][1]=edge[i].color;
        }
    }while (head<tail);
}
void update(int k,int from)
{
    g[deep[k]]=max(g[deep[k]],len[k]);
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from&&!flag[edge[i].to]) update(edge[i].to,k);
}
void solve(int k)
{
    make(k,k);k=findroot(k,size[k]);
    flag[k]=1;deep[k]=0;make(k,k);
    int cnt=0;
    for (int i=p[k];i;i=edge[i].nxt)
    if (!flag[edge[i].to])
    {
        colordeep[edge[i].color]=mxdeep[edge[i].to]=0,findmx(edge[i].to,edge[i].to,edge[i].to);
        cnt++,v[cnt].x=edge[i].to,v[cnt].y=edge[i].color;
    }
    for (int i=p[k];i;i=edge[i].nxt)
    if (!flag[edge[i].to]) colordeep[edge[i].color]=max(colordeep[edge[i].color],mxdeep[edge[i].to]);
    sort(v+1,v+cnt+1);
    for (int i=1;i<=cnt;i++)
    {
        int t=i-1,d=0;
        while (t1].y==v[i].y)
        {
            len[v[++t].x]=a[v[i].y];
            bfs(v[t].x,v[t].y,-a[v[i].y],d);
            update(v[t].x,v[t].x);d=mxdeep[v[t].x];
        }
        int H=1,T=0;for (int j=min(colordeep[v[i-1].y],r);j>=l;j--) ins(j,H,T,f);
        for (int j=1;j<=d;j++)
        {
            pop(r-j,H,T);
            if (l>=j&&l-j<=d) ins(l-j,H,T,f);
            if (H<=T) ans=max(ans,g[j]+f[Q[H]]);
        }
        for (int j=0;j<=d;j++) f[j]=max(f[j],g[j]),g[j]=-inf;
        i=t;
    }
    for (int i=0;i<=mxdeep[v[cnt].x];i++) f[i]=-inf;
    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("bzoj4860.in","r",stdin);
    freopen("bzoj4860.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read(),l=read(),r=read();
    for (int i=0;i<=n;i++) f[i]=g[i]=-inf;
    for (int i=1;i<=m;i++) a[i]=read();
    for (int i=1;i)
    {
        int x=read(),y=read(),z=read();
        addedge(x,y,z),addedge(y,x,z);
    }
    solve(1);
    cout<<ans;
    return 0;
}

 

转载于:https://www.cnblogs.com/Gloid/p/10004006.html