bzoj3600 没有人的算术

查询可以直接线段树维护，修改呢，考虑一颗替罪羊，每个点代表一段区间，他的$val$就是区间中值，线段树记录对应节点的$id$，再开一个数组记录即时的权值，因为重建时$val$可能会变。这好像是重量平衡树的应用，然而究竟什么是重量平衡树呢？

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<cmath>
  6 #define inf 0x3fffffffffffffff
  7 #define alp 0.75
  8 #define N 500050
  9 using namespace std;
 10 double a[500050];
 11 int tot;
 12 struct Node{
 13     Node *ch[2];
 14     double l,r;
 15     int size,cover,ex,k1,k2,id;
 16     void pushup(){
 17         size=ch[0]->size+ch[1]->size+ex;
 18         cover=ch[0]->cover+ch[1]->cover+1;
 19     }
 20     bool bad(){
 21         return ch[0]->cover>=cover*alp+5||ch[1]->cover>=cover*alp+5;
 22     }
 23     Node(double x,double y,int z,int w,int k);
 24 }*null=new Node(0,0,0,0,0),*root,*sta[N];
 25 int len;
 26 Node:: Node(double x,double y,int z,int w,int k){
 27     l=x;r=y;k1=z;k2=w;id=k;
 28     a[k]=(l+r)/2.0;
 29     size=cover=ex=1;
 30     ch[0]=ch[1]=null;
 31 }
 32 void init(){
 33     null->ch[0]=null->ch[1]=null;
 34     null->l=null->r=null->k1=null->k2=null->id=0;
 35     null->size=null->cover=null->ex=0;
 36     a[0]=-5000000000000000000;
 37     root=new Node(-inf,inf,0,0,++tot);
 38 }
 39 Node **insert(Node *&rt,double l,double r,int k1,int k2){
 40     if(rt==null){
 41         rt=new Node(l,r,k1,k2,++tot);
 42         return &null;
 43     }
 44     rt->size++;rt->cover++;
 45     Node **ret;
 46     if(a[rt->k1]<a[k1]||(a[rt->k1]==a[k1]&&a[rt->k2]<a[k2]))ret=insert(rt->ch[1],(rt->l+rt->r)/2.0,rt->r,k1,k2);
 47     else ret=insert(rt->ch[0],rt->l,(rt->l+rt->r)/2.0,k1,k2);
 48     if(rt->bad())ret=&rt;
 49     return ret;
 50 }
 51 void travel(Node *rt){
 52     if(rt==null)return;
 53     travel(rt->ch[0]);
 54     if(rt->ex){
 55         sta[++len]=rt;
 56         travel(rt->ch[1]);
 57     }
 58     else{
 59         travel(rt->ch[1]);
 60         delete rt;
 61     }
 62 }
 63 Node * divide(int l,int r,double L,double R){
 64     if(l>r)return null;
 65     int mid=(l+r)>>1;
 66     sta[mid]->l=L;sta[mid]->r=R;a[sta[mid]->id]=(L+R)/2.0;
 67     sta[mid]->ch[0]=divide(l,mid-1,L,(L+R)/2.0);
 68     sta[mid]->ch[1]=divide(mid+1,r,(L+R)/2.0,R);
 69     sta[mid]->pushup();
 70     return sta[mid];
 71 }
 72 void rebuild(Node *&rt){
 73     double L=rt->l,R=rt->r;
 74     len=0;travel(rt);
 75     rt=divide(1,len,L,R);
 76 }
 77 void insert(int k1,int k2){
 78     Node **p=insert(root,-inf,inf,k1,k2);
 79     if(*p!=null)rebuild(*p);
 80 }
 81 Node *find(Node *rt,int k1,int k2){
 82     if(rt==null)return rt;
 83     if(rt->ex&&a[rt->k1]==a[k1]&&a[rt->k2]==a[k2])return rt;
 84     if(a[rt->k1]<a[k1]||(a[rt->k1]==a[k1]&&a[rt->k2]<a[k2]))return find(rt->ch[1],k1,k2);
 85     else return find(rt->ch[0],k1,k2);
 86 }
 87 
 88 int w[N],maxn[N],maxpos[N];
 89 void pushup(int rt){
 90     if(a[maxn[rt<<1]]>=a[maxn[rt<<1|1]])
 91         maxn[rt]=maxn[rt<<1],
 92         maxpos[rt]=maxpos[rt<<1];
 93     else maxn[rt]=maxn[rt<<1|1],
 94         maxpos[rt]=maxpos[rt<<1|1];
 95 }
 96 void build(int rt,int l,int r){
 97     if(l==r){maxn[rt]=w[l];maxpos[rt]=l;return ;}
 98     int mid=(l+r)>>1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);
 99     pushup(rt);
100 }
101 void update(int rt,int l,int r,int x,int y){
102     if(l==r){maxn[rt]=y;maxpos[rt]=l;return ;}
103     int mid=(l+r)>>1;
104     if(x<=mid)update(rt<<1,l,mid,x,y);
105     else update(rt<<1|1,mid+1,r,x,y);
106     pushup(rt);
107 }
108 int MX,POS;
109 void query(int rt,int l,int r,int x,int y){
110     if(x<=l&&r<=y){
111         if(a[MX]<a[maxn[rt]])
112             MX=maxn[rt],POS=maxpos[rt];
113         return;
114     }
115     int mid=(l+r)>>1;
116     if(x<=mid)query(rt<<1,l,mid,x,y);
117     if(y>mid)query(rt<<1|1,mid+1,r,x,y);
118 }
119 
120 int n,m;
121 int main(){
122     init();
123     scanf("%d%d",&n,&m);
124     for(int i=1;i<=n;i++)w[i]=1;
125     build(1,1,n);
126     char ch[2];
127     int x,y,z,l,r;
128     while(m--){
129         scanf("%s",ch);
130         if(ch[0]=='C'){
131             scanf("%d%d%d",&x,&y,&z);
132             Node *now=find(root,w[x],w[y]);
133             if(now!=null)w[z]=now->id;
134             else{insert(w[x],w[y]);w[z]=tot;}
135             update(1,1,n,z,w[z]);
136         }
137         if(ch[0]=='Q'){
138             scanf("%d%d",&l,&r);
139             MX=POS=0;query(1,1,n,l,r);
140             printf("%d\n",POS);
141         }
142     }
143     return 0;
144 }
View Code

 

posted @ 2018-01-15 13:25  Ren_Ivan  阅读(245)  评论(0)  编辑  收藏  举报