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  • 【BZOJ3545】 [ONTAK2010]Peaks

    BZOJ3545 [ONTAK2010]Peaks


    Solution

    既然会加强版,直接把强制在线的操作去掉就好了.

    代码实现

    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<algorithm>
    #include<queue>
    #include<set>
    #include<map>
    #include<iostream>
    using namespace std;
    #define ll long long
    #define re register
    #define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
    inline int gi(){
    	int f=1,sum=0;char ch=getchar();
    	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    	return f*sum;
    }
    int two[20];
    int n,m,Q,cnt,top,tot,nodes,lastans;
    int H[100005],h[100005];
    int f[200005],val[200005],st[200005],ed[200005];
    int q[300005],rt[300005];
    int ls[5000005],rs[5000005],sum[5000005];
    int dep[200005],fa[200005][17],mx[200005][17];
    bool vis[200005];
    struct node{int u,v,w;}e[500005];
    int to[200005],nxt[200005],front[200005];
    bool operator<(node a,node b){
    	return a.w<b.w;
    }
    inline void Add(int u,int v){
    	to[++cnt]=v;nxt[cnt]=front[u];front[u]=cnt;
    }
    inline int find(int x){
    	if(f[x]!=x)f[x]=find(f[x]);
    	return f[x];
    }
    inline int Find(int x,int val){
    	for(int i=17;~i;i--)
    		if(dep[x]>=two[i] && mx[x][i]<=val)x=fa[x][i];
    	return x;
    }
    inline void insert(int l,int r,int &x,int f,int val){
    	x=++tot;
    	sum[x]=sum[f]+1;
    	if(l==r)return;
    	ls[x]=ls[f];rs[x]=rs[f];
    	int mid=(l+r)>>1;
    	if(val<=mid)insert(l,mid,ls[x],ls[f],val);
    	else insert(mid+1,r,rs[x],rs[f],val);
    }
    inline int query(int l,int r,int x,int y,int rank){
    	if(l==r)return l;
    	int mid=(l+r)>>1;
    	if(sum[ls[y]]-sum[ls[x]]>=rank)return query(l,mid,ls[x],ls[y],rank);
    	else return query(mid+1,r,rs[x],rs[y],rank-sum[ls[y]]+sum[ls[x]]);
    }
    inline void dfs(int u){
    	vis[u]=1;q[++top]=u;
    	for(int i=1;i<=16;i++)
    		if(dep[u]>=two[i]){
    			fa[u][i]=fa[fa[u][i-1]][i-1];
    			mx[u][i]=max(mx[u][i-1],mx[fa[u][i-1]][i-1]);
    		}
    		else break;
    	for(int i=front[u];i;i=nxt[i]){
    		int v=to[i];
    		dep[v]=dep[u]+1;
    		mx[v][0]=val[u];
    		fa[v][0]=u;
    		dfs(v);
    	}
    	if(u>n)q[++top]=u;
    }
    inline void solve(){
    	for(re int i=1;i<=Q;i++){
    		re int v=gi(),x=gi(),k=gi();
    //		if(lastans!=-1)v^=lastans,x^=lastans,k^=lastans;
    		re int t=Find(v,x);
    		re int a=rt[st[t]],b=rt[ed[t]];
    		if(sum[b]-sum[a]<k)lastans=-1;
    		else lastans=H[query(1,n,a,b,sum[b]-sum[a]-k+1)];
    		printf("%d
    ",lastans);
    	}
    }
    int main()
    {
    	two[0]=1;
    	for(re int i=1;i<20;i++)two[i]=two[i-1]<<1;
    	n=gi();m=gi();Q=gi();
    	for(re int i=1;i<=n;i++)h[i]=gi(),H[i]=h[i];
    	sort(H+1,H+n+1);
    	for(re int i=1;i<=n;i++)h[i]=lower_bound(H+1,H+n+1,h[i])-H;
    	for(re int i=1;i<=n+n;i++)f[i]=i;
    	for(re int i=1;i<=m;i++)
    		e[i].u=gi(),e[i].v=gi(),e[i].w=gi();
    	sort(e+1,e+m+1);
    	nodes=n;
    	for(re int i=1;i<=m;i++){
    		re int u=find(e[i].u),v=find(e[i].v);
    		if(u!=v){
    			nodes++;
    			f[u]=f[v]=nodes;
    			val[nodes]=e[i].w;
    			Add(nodes,u);Add(nodes,v);
    			if(nodes==2*n-1)break;
    		}
    	}
    	for(re int i=1;i<=n;i++)
    		if(!vis[i])dfs(find(i));
    	for(re int i=1;i<=top;i++){
    		re int t=q[i];
    		if(t<=n)insert(1,n,rt[i],rt[i-1],h[t]);
    		else{
    			rt[i]=rt[i-1];
    			if(!st[t])st[t]=i;
    			else ed[t]=i;
    		}
    	}
    	solve();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mleautomaton/p/10315861.html
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