Hdu5412 -- CRB and Queries (binary classification)

Hdu5412 -- CRB and Queries (binary classification)

 

Given the initial sequence of n numbers, there are two operations: 1 l v switches the number of l to v, 2 l r k asks in the interval [l, r] and Output

Classic questions, but the tree array + the Chair tree (TLE) stretch tree (MLE), I heard that they use the block chain list, zhazha said no, then I will make up the question, it is a good method to discover the entire binary classification.

First of all, this entire binary is to put the data range and operation together, and continue to divide the data range. Some operations can be completed, and some operations cannot be completed.

For example, if the range of the current data is [l, r], then mid = (l + r)/2. If the number of modifications is less than or equal to mid, the number of modifications can be completed. Otherwise, the modification cannot be completed. For a query operation, if the number x in the queried range is smaller than k, the number to be queried is not in the range [1, 5, therefore, this operation will not be completed in [l, mid], and the k of this operation will be changed to k-x. Otherwise, it will be completed in [l, mid. [L, mid], [mid + 1, r] are obtained based on the data range. operations that can be completed are placed in the left interval. operations that cannot be completed are placed in the right interval. (Note that the relative sequence of operations placed in the same interval cannot be changed.) in this way, the final query operation will be determined to a value, which is the value to be output.

 

#include 
 
  #include 
  
   #include using namespace std ;#define maxn 300100struct node{    int type, l, r, v , ans;}p[maxn];int c[maxn], n, q, cnt, max1;int a[maxn];int id1[maxn], id2[maxn];void p_add(int type, int l,int r, int v) {    p[cnt].type = type;    p[cnt].l = l; p[cnt].r = r;    p[cnt].v = v;    id1[cnt] = cnt;    cnt++;}int lowbit(int x) {    return x & -x ;}void add(int i,int k) {    while( i <= n ) {        c[i] += k ;        i += lowbit(i) ;    }}int sum(int i) {    int num = 0 ;    while( i ) {        num += c[i] ;        i -= lowbit(i) ;    }    return num ;}void solve(int L, int R, int low, int high) {    if( L > R ) return;    if( low == high ) {        while(L <= R) {            if( p[ id1[L] ].type == 2 ) p[ id1[L] ].ans = low;            L++;        }        return ;    }    int i, j, num, l = L, r = R;    int mid = (low + high)/2;    for(i = L; i <= R; i++) {        j = id1[i];        if( p[j].type == 2 ) {            num = sum(p[j].r) - sum(p[j].l-1);            if( num < p[j].v ) {                p[j].v -= num;                id2[r--] = j;            }            else                id2[l++] = j;        }        else {            if( p[j].v <= mid ) {                add(p[j].l,p[j].type);                id2[l++] = j;            }            else                id2[r--] = j;        }    }    for(i = L; i <= R; i++) {        j = id1[i];        if( p[j].type != 2 && p[j].v <= mid ) add(p[j].l,-p[j].type);    }    for(i = L; i < l; i++)        id1[i] = id2[i];    for(r = R; i <= R; r--, i++)        id1[i] = id2[r];    solve(L,l-1,low,mid);    solve(l,R,mid+1,high);}int main() {    int i, j, type, l, r, v;    while( scanf(%d, &n) != EOF ) {        memset(c,0,sizeof(c));        cnt = max1 = 0;        for(i = 1; i <= n; i++) {            scanf(%d, &a[i]);            p_add(1, i, i, a[i]);            max1 = max(max1, a[i]);        }        scanf(%d, &q);        while( q-- ) {            scanf(%d, &type);            if( type == 1 ) {                scanf(%d %d, &l, &v);                p_add(-1, l, l, a[l]);                a[l] = v;                p_add(1, l, l, a[l]);                max1 = max(max1, a[l]);            }            else {                scanf(%d %d %d, &l, &r, &v);                p_add(2, l, r, v);            }        }        solve(0,cnt-1,0,max1) ;        for(i = 0; i < cnt; i++) {            if( p[i].type == 2 )                printf(%d, p[i].ans);        }    }    return 0 ;}