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  • bzoj2120 数颜色 分块

    分块大法好 orz

    处理出每个点的前驱和后继位置。

    暴力修改,查询就在每个整块里查询pre<l的,暴力跑两边就好了

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #define N 10005
    using namespace std;
    int n,m,nn,a[N],be[N],pre[N],nxt[N];
    int last[1000005],pp[N],ans;
    void work(int x){
    	int l=(x-1)*nn+1,r=min(n,x*nn);
    	for(int i=l;i<=r;i++)
    		pp[i]=pre[i];
    	sort(pp+l,pp+r+1);
    }
    void change(int x,int y){
    	a[x]=y;
    	if(nxt[x]){pre[nxt[x]]=pre[x];work(be[nxt[x]]);}
    	if(pre[x]){nxt[pre[x]]=nxt[x];work(be[pre[x]]);}
    	int pr=0,ne=0;
    	for(int i=1;i<=n;i++){
    		if(i<x&&a[i]==y) pr=i;
    		if(i>x&&a[i]==y){ne=i;break;}
    	}
    	pre[x]=pr; nxt[x]=ne;
    	work(be[x]); 
    	 if(pr) {nxt[pr]=x;work(be[pr]);}
    	 if(ne) {pre[ne]=x;work(be[ne]);}
    }
    int main()
    {
    	scanf("%d%d",&n,&m);
    	nn=(int) sqrt(n);
    	for(int i=1;i<=n;i++){
    		scanf("%d",&a[i]);
    		be[i]=(i-1)/nn+1;
    	}
    	for(int i=1;i<=n;i++){
    		pre[i]=last[a[i]];
    		last[a[i]]=i;
    		if(pre[i]) nxt[pre[i]]=i;
    	}
    	int tot=be[n];
    	for(int i=1;i<=tot;i++)work(i);
    	char ch; int x,y,l,r,num;
    	while(m--){
    		ch=getchar();
    		while(ch!='Q'&&ch!='R')ch=getchar();
    		scanf("%d%d",&x,&y);
    		if(ch=='Q'){
    			ans=0;
    			if(be[x]==be[y]){
    				for(int i=x;i<=y;i++)if(pre[i]<x)ans++;
    				printf("%d
    ",ans); continue;
    			}
    			for(int i=x;i<=be[x]*nn;i++)
    				if(pre[i]<x) ans++;
    			for(int i=(be[y]-1)*nn+1;i<=y;i++)
    				if(pre[i]<x) ans++;
    			for(int i=be[x]+1;i<be[y];i++){
    				l=(i-1)*nn+1,r=min(n,i*nn); 
    				num=lower_bound(pp+l,pp+r+1,x)-pp;
    				ans+=num-l;
    			}
    			printf("%d
    ",ans);
    		}
    		if(ch=='R') change(x,y);
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/Ren-Ivan/p/7746750.html
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