bzoj 1901 动态区间第K大 树状数组套可持久化数据结构

空间复杂度 nlgnlgn  在zojMLE  大视野 AC

第一次用树套树。。写的 不优雅。

/**************************************************************    Problem: 1901    User: OceanLight    Language: C++    Result: Accepted    Time:444 ms    Memory:12416 kb****************************************************************/ #include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <cstring>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <assert.h>#include <queue>#define REP(i,n) for(int i=0;i<n;i++)#define TR(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)#define ALLL(x) x.begin(),x.end()#define SORT(x) sort(ALLL(x))#define CLEAR(x) memset(x,0,sizeof(x))#define FILLL(x,c) memset(x,c,sizeof(x))using namespace std;const double eps = 1e-9;#define LL long long#define pb push_backconst int maxn  = 11000;int num[maxn];int n , m ;map<int ,int>mp;map<int,int>::iterator it;char op[maxn][4];int t[maxn][3];int tot; struct Node {    Node *l,*r;    int sum;}nodes[maxn*80];Node *root[maxn];Node *null;int C;void init(){    C= 0;    null = &nodes[C++];    null->l = null->r = null;    null->sum =0;    root[0] = null;    for(int i=1;i<=n;i++){         root[i] = &nodes[C++];        root[i]->l = null;        root[i]->r = null;        root[i]->sum  = 0;    }}Node * update(int pos,int left,int right,Node *root,int val){     //cout << pos << "  "<<left << "  "<< right <<  "     "<<val<<endl;;      Node * rt = root;      if(rt == null){           rt = &nodes[C++];           rt->l = null;           rt->r = null;           rt->sum = 0;      }      if(left == right){            rt->sum +=val;            return rt;      }      int mid = (left + right)/2;      if(pos<=mid){         rt->l = update(pos,left,mid,rt->l,val);      }else{          rt->r = update(pos,mid+1,right,rt->r ,val);      }      rt->sum = rt->l->sum + rt->r->sum ;      //cout << pos << "  "<<left << "  "<< right <<  "     "<<val<<"  "<< rt->sum <<endl;      return rt;} #define lowbit(k)   k & -kvoid inc(int k ,int m,int add) {    //k位置上，加入值m   // m位置上的数+= add        for(; k <= n; k += lowbit(k))           root[k] = update(m,1,tot,root[k],add);}int l_num,r_num;Node *ll[maxn],*rr[maxn];void get(int k,bool left){       int idx = 0;        for(; k > 0; k -= lowbit(k)){            idx++ ;            if(left){                 ll[idx] = root[k];            }else{                rr[idx] = root[k];            }        }        if(left){            l_num = idx;        }else{            r_num = idx;        }}int query(int left,int right,int k){       if(left==right){           return left;       }       int l_sum = 0;       for(int i=1;i<=l_num;i++){             l_sum += ll[i]->l->sum;       }       int r_sum = 0;       for(int i =1;i<=r_num;i++){             r_sum += rr[i]->l->sum;       }       int a = r_sum - l_sum;      //cout << a <<endl;       int mid = (left +right)/2;       if(a>=k){            for(int i=1;i<=l_num ;i++){                ll[i] = ll[i]->l;            }            for(int i=1;i<=r_num;i++){                 rr[i] = rr[i]->l;            }            return query(left,mid,k);       }else{            for(int i=1;i<=l_num ;i++){                ll[i] = ll[i]->r;            }            for(int i=1;i<=r_num;i++){                 rr[i] = rr[i]->r;            }            return query(mid+1,right,k-a);       }      }int idx[maxn*2];void solve(){    init();    for(int i=1;i<=n;i++){        inc(i,mp[num[i]],1);    }    //cout << C<<endl;    for(int i=1;i<=m;i++){        if(op[i][0]== 'Q'){            int left = t[i][0];            int right = t[i][1];            int k = t[i][2];            get(left-1,1);            get(right,0);            int ans = query(1,tot,k);            printf("%d\n",idx[ans]);        }else{            int idx = t[i][0];            int val= t[i][1];            inc(idx,mp[num[idx]],-1);            num[idx] = val;            inc(idx,mp[num[idx]],1);        }    }     } int main(){    while(~scanf("%d%d",&n,&m)){        mp.clear();        for(int i=1;i<=n;i++){            scanf("%d",&num[i]);            mp[num[i]]=1;        }        for(int i=1;i<=m;i++){            scanf("%s",op[i]);            if(op[i][0]== 'Q'){                 scanf("%d%d%d",&t[i][0],&t[i][1],&t[i][2]);            }else{                scanf("%d%d",&t[i][0],&t[i][1]);                mp[t[i][1]]=1;            }        }        tot= 0 ;        for(it = mp.begin(); it!= mp.end();it++){            tot ++ ;             it->second = tot;             idx[tot] = it->first;        }        solve();    }      return 0;}