zoj 2112(主席树套树状数组+优化)

题解思路:如果直接对原来的数组建立主席树套树状数组的空间复杂度是（（n+m）log(n+m)*log(n+m)）这样明显爆炸，那么我们可以对原来的数组建主席树模型，空间复杂度就是n*logn，对m里面的修改建主席树套树状数组空间复杂度是m*logm*logm明显小了很多。

#include<bits/stdc++.h>#define lson l,mid#define rson mid+1,rusing namespace std;typedef long long ll;int n,m,k,cnt,R,L;const int mx = 6e4+10;int num[mx],root[mx],kep[mx],ran[mx],now[mx];int ls[30*mx],rs[30*mx],sum[30*mx],size;struct node{int flag,val;int l,r;}s[mx/3];void update(int x,int &y,int l,int r,int M,int v){y = ++size;    ls[y] = ls[x],rs[y] = rs[x],sum[y] = sum[x]+v;    int mid = (l+r)>>1;    if(l==r) return ;//更新一边的儿子    if(M<=mid) update(ls[x],ls[y],lson,M,v);    else update(rs[x],rs[y],rson,M,v);}inline int lowbit(int x){  return x&(-x);  }int get_sum(int x){int ans = 0;while(x){ans += sum[ls[kep[x]]];x -= lowbit(x);}return ans;}int query(int l,int r,int lt,int rt,int k){if(l==r) return l;int mid = (l+r)>>1;          //修改后变化的+原来的差 int ret = get_sum(R)-get_sum(L)+sum[ls[rt]]-sum[ls[lt]];    if(k<=ret){    for(int i=L;i>0;i-=lowbit(i)) kep[i] = ls[kep[i]];//跟踪儿子结点     for(int i=R;i>0;i-=lowbit(i)) kep[i] = ls[kep[i]];     return query(lson,ls[lt],ls[rt],k);}    for(int i=L;i>0;i-=lowbit(i)) kep[i] = rs[kep[i]];    for(int i=R;i>0;i-=lowbit(i)) kep[i] = rs[kep[i]];    return query(rson,rs[lt],rs[rt],k-ret);}void add(int x,int p,int v){for(int i=x;i<=cnt;i+=lowbit(i))update(now[i],now[i],1,n,p,v);}int main(){    char str[10];    //freopen("1in","r",stdin);    int t;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        memset(sum,0,sizeof(sum));        memset(now,0,sizeof(now));        ls[0] = rs[0] = size = 0;        for(int i=1;i<=n;i++) scanf("%d",num+i),ran[i] = num[i];        k = cnt = n;        for(int i=1;i<=m;i++){        scanf("%s",str);        if(str[0]=='Q'){        s[i].flag = 1;        scanf("%d%d%d",&s[i].l,&s[i].r,&s[i].val);    }else{    s[i].flag = 0;    scanf("%d%d",&s[i].l,&s[i].val);    ran[++k] = s[i].val;      }    }    n = 1;    sort(ran+1,ran+k+1);    for(int i=2;i<=k;i++) if(ran[i]!=ran[i-1]) ran[++n] = ran[i];for(int i=1;i<=cnt;i++){            int p = lower_bound(ran+1,ran+1+n,num[i]) - ran;            update(root[i-1],root[i],1,n,p,1);        }  for(int i=1;i<=m;i++){if(s[i].flag){L = --s[i].l;R = s[i].r;for(int j=L;j>0;j-=lowbit(j)) kep[j] = now[j];for(int j=R;j>0;j-=lowbit(j)) kep[j] = now[j];printf("%d\n",ran[query(1,n,root[L],root[R],s[i].val)]);}else{int p = lower_bound(ran+1,ran+1+n,num[s[i].l])-ran;add(s[i].l,p,-1);p = lower_bound(ran+1,ran+1+n,s[i].val)-ran;add(s[i].l,p,1); num[s[i].l] = s[i].val;}}}    return 0;}