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  • hdu 3804 树链剖分

    思路:将边权排序,然后插入线段树,这样就可以直接用二分查找确定答案。

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<set>
    #include<map>
    #include<cmath>
    #include<queue>
    #include<cstdio>
    #include<vector>
    #include<string>
    #include<cstdlib>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define pb push_back
    #define mp make_pair
    #define Maxn 100010
    #define Maxm 200010
    #define LL __int64
    #define Abs(x) ((x)>0?(x):(-x))
    #define lson(x) (x<<1)
    #define rson(x) (x<<1|1)
    #define inf 100000
    #define lowbit(x) (x&(-x))
    #define clr(x,y) memset(x,y,sizeof(x))
    #define Mod 1000000007
    using namespace std;
    int head[Maxn],vi[Maxn],dep[Maxn],w[Maxn],top[Maxn],son[Maxn],sz[Maxn],fa[Maxn],e,id;
    int val[Maxn];
    int num[Maxn];
    struct Point{
        int val,i;
        int operator <(const Point &temp) const{
            return val<temp.val;
        }
    }lis[Maxn];
    struct Edge{
        int u,v,next,val;
    }edge[Maxn*3];
    struct Tree{
        int l,r,c;
        int *p;
        int mid(){
            return (l+r)>>1;
        }
    }tree[Maxn*3];
    void init()
    {
        clr(head,-1);clr(vi,0);
        e=0;id=0;
        for(int i=0;i<Maxn*3;i++){
            delete [] tree[i].p;
        }
    }
    void add(int u,int v,int val)
    {
        edge[e].u=u,edge[e].v=v,edge[e].val=val,edge[e].next=head[u],head[u]=e++;
    }
    void BuildTree(int l,int r,int po)
    {
        tree[po].l=l,tree[po].r=r,tree[po].c=0;
        tree[po].p=new int[r-l+2];
        if(l==r)
            return ;
        int mid=tree[po].mid();
        BuildTree(l,mid,lson(po));
        BuildTree(mid+1,r,rson(po));
    }
    void update(int i,int c,int po)
    {
        if(tree[po].l==tree[po].r){
            tree[po].p[++tree[po].c]=c;
            return ;
        }
        tree[po].p[++tree[po].c]=c;
        int mid=tree[po].mid();
        if(i<=mid)
            update(i,c,lson(po));
        else
            update(i,c,rson(po));
    }
    int getans(int l,int r,int val,int po)
    {
        if(l<=tree[po].l&&tree[po].r<=r){
            int pos=lower_bound(tree[po].p+1,tree[po].p+tree[po].c+1,val)-tree[po].p;
           // cout<<tree[po].l<<" "<<tree[po].r<<" "<<l<<" "<<r
            if(pos>tree[po].c)
                return tree[po].p[tree[po].c];
            if(tree[po].p[pos]==val)
                return val;
            if(pos>1){
                return tree[po].p[pos-1];
            }
            return -1;
        }
        int mid=tree[po].mid();
        if(r<=mid)
            return getans(l,r,val,lson(po));
        else if(l>=mid+1)
            return getans(l,r,val,rson(po));
        else{
            return max(getans(l,mid,val,lson(po)),getans(mid+1,r,val,rson(po)));
        }
    }
    void dfs(int u,int val)
    {
        vi[u]=1;
        int i,v;
        son[u]=0,sz[u]=1;
        num[u]=val;
        for(i=head[u];i!=-1;i=edge[i].next){
            v=edge[i].v;
            if(vi[v]) continue;
            dep[v]=dep[u]+1;
            fa[v]=u;
            dfs(v,edge[i].val);
            if(sz[v]>sz[son[u]])son[u]=v;
            sz[u]+=sz[v];
        }
    }
    void build(int u,int ti)
    {
        int i,v;
        w[u]=++id;top[u]=ti;vi[u]=1;
        lis[id].i=id,lis[id].val=num[u];
        if(son[u]) build(son[u],ti);
        for(i=head[u];i!=-1;i=edge[i].next){
            v=edge[i].v;
            if(vi[v]||v==son[u]) continue;
            build(v,v);
        }
    }
    void calc(int u,int c)
    {
        int f1=top[u];
        int ans=-10000;
        while(f1!=1){
            ans=max(ans,getans(w[f1],w[u],c,1));
            u=fa[f1];f1=top[u];
        }
        //cout<<w[f1]<<" "<<w[u]<<endl;
        ans=max(ans,getans(w[f1],w[u],c,1));
        printf("%d
    ",ans);
        return ;
    }
    int main()
    {
        int t,n,i,j,u,v,val,q;
        scanf("%d",&t);
        while(t--){
            init();
            scanf("%d",&n);
            for(i=1;i<n;i++){
                scanf("%d%d%d",&u,&v,&val);
                add(u,v,val);
                add(v,u,val);
            }
            dfs(1,1200000000);
            memset(vi,0,sizeof(vi));
            build(1,1);
            sort(lis+1,lis+id+1);
            BuildTree(1,n,1);
            for(i=1;i<=n;i++){
                update(lis[i].i,lis[i].val,1);
            }
            scanf("%d",&q);
            for(i=1;i<=q;i++){
                scanf("%d%d",&u,&v);
                calc(u,v);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wangfang20/p/3321001.html
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