POJ-2104 K-th Number CDQ分治

题目传送门

题意：给你一个序列，长度为n，m次询问，询问一段区间的第k大。

题解：CDQ分治，对整个值域进行分治。每次取一个mid， 计算出整个区间内mid <= 的数目，如果 num >= k， 那么就可以第k大的数一定落在[ l, mid]之间， 否者就会落在 [mid+1, r]之间， 然后我们继续递归二分下去，直到 l == r 那么就找到第k大了。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<iostream>
 #include<cstring>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = 2e5 + ;
 int n;
 struct Node{
     int op;
     int L, R;
     int x, id;
 }A[N], lf[N], ri[N];
 int tot = ;
 void NowNode(int op, int L, int R, int x, int id){
     ++tot;
     A[tot].op = op; A[tot].L = L; A[tot].R = R;
     A[tot].x = x; A[tot].id = id;
 }
 int ans[N];
 int tree[N];
 int Mn = inf, Mx = ;
 void Add(int x, int v){
     for(int i = x; i <= n; i += i & (-i)){
         tree[i] += v;
     }
 }
 int Query(int x){
     int ret = ;
     for(int i = x; i; i -= i&(-i)){
         ret += tree[i];
     }
     return ret;
 }
 void cdq(int Lval, int Rval, int st, int ed){
     if(Lval > Rval || st > ed) return ;
     if(Lval == Rval){
         for(int i = st; i <= ed; i++)
             if(A[i].op == ) ans[A[i].id] = Lval;
         return ;
     }
     int mid = Lval + Rval >> ;
 
     int lsz = , rsz = ;
     for(int i = st; i <= ed; i++){
         int op = A[i].op, x = A[i].x, L = A[i].L, R = A[i].R, id = A[i].id;
         if(op == ){
             if(x <= mid){
                 Add(A[i].id, );
                 lf[++lsz] = A[i];
             }
             else ri[++rsz] = A[i];
         }
         else if(A[i].op == ){
             int num = Query(R) - Query(L-);
             if(num >= x) lf[++lsz] = A[i];
             else {
                 A[i].x -= num;
                 ri[++rsz] = A[i];
             }
         }
     }
     for(int i = ; i <= lsz; i++) {
 
         if(lf[i].op == ) {
             Add(lf[i].id, -);
         }
         A[st + i - ] = lf[i];
     }
     for(int i = ; i <= rsz; i++) A[st + lsz + i - ] = ri[i];
     cdq(Lval, mid, st, st+lsz-);
     cdq(mid+, Rval, st+lsz, ed);
 }
 int main(){
     int m, val, l, r;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; i++){
         scanf("%d", &val);
         NowNode(,,,val,i);
         Mn = min(Mn, val);
         Mx = max(Mx, val);
     }
     for(int i = ; i <= m; i++){
         scanf("%d%d%d", &l, &r, &val);
         NowNode(,l,r,val,i);
     }
     cdq(Mn,Mx,,n+m);
     for(int i = ; i <= m; i++){
         printf("%d\n", ans[i]);
     }
     return ;
 }