【jzoj5221】【GDOI2018模拟7.10】【A】【线段树合并】

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题目大意

解题思路

从下往上建权值线段树，用子树的线段树合并出当前的线段树，维护最大连续区间和size即可。

code

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define LF double#define LL long long#define ULL unsigned int#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)#define fr(i,j) for(int i=begin[j];i;i=next[i])using namespace std;int max(int x,int y){return (x>y)?x:y;}int min(int x,int y){return (x<y)?x:y;}int const mn=1e5+9,mp=18*1e5+9,inf=1e9+7;int n,gra,pon,ans,begin[mn],to[mn],next[mn],num[mn],pre[mn],mx[mp],lmx[mp],    rmx[mp],size[mp],son[mp][2];void insert(int u,int v){    to[++gra]=v;    next[gra]=begin[u];    begin[u]=gra;}void update(int p,int l,int r,int md){    int ls=son[p][0],rs=son[p][1];    mx[p]=max(max(mx[ls],mx[rs]),rmx[ls]+lmx[rs]);    lmx[p]=(lmx[ls]==md-l+1)?md-l+1+lmx[rs]:lmx[ls];    rmx[p]=(rmx[rs]==r-md)?r-md+rmx[ls]:rmx[rs];    size[p]=size[ls]+size[rs];}void oper(int p,int l,int r,int v){    int md=(l+r)/2;    if(l==r){        mx[p]=lmx[p]=rmx[p]=size[p]=1;        return;    }    if(v<=md){        if(!son[p][0])son[p][0]=++pon;        oper(son[p][0],l,md,v);    }else{        if(!son[p][1])son[p][1]=++pon;        oper(son[p][1],md+1,r,v);    }    update(p,l,r,md);}int merge(int p,int q,int l,int r){    if(size[p]<size[q])swap(p,q);    if(!size[q])return p;    int md=(l+r)/2;    son[p][0]=merge(son[p][0],son[q][0],l,md);    son[p][1]=merge(son[p][1],son[q][1],md+1,r);    update(p,l,r,md);    return p;}void dfs(int now){    num[now]=++pon;    oper(pon,1,n,now);    fr(i,now){        dfs(to[i]);        num[now]=merge(num[now],num[to[i]],1,n);    }    if((mx[num[now]]>5)&&(mx[num[now]]!=n)&&(mx[num[now]]==size[num[now]])){        int bb;        bb++;    }    ans+=mx[num[now]]==size[num[now]];}int main(){    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d",&n);    fo(i,1,n-1){        int u,v;        scanf("%d%d",&u,&v);        insert(u,v);        pre[v]++;    }    int rt;    fo(i,1,n)if(!pre[i]){rt=i;break;}    dfs(rt);    printf("%d",ans);    return 0;}

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