hdu 3971

 

 杭电2015级新生如何加入ACM集训队？ 

Play With Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 452    Accepted Submission(s): 228

Problem Description

When the girl was solving GSSX, a serious of tough problems about data structure on SPOJ, something intriguing once again comes to GYZ's mind. That is, for a changing sequences, how to count how many elements in a specific range efficiently. 

Without any beneficial idea, as usual, GYZ asks her friend, CLJ for help. But this time, unfortunately, CLJ is playing a gal-game at present, does not have sparse time. 

So now , it is your turn... 

Cause the original problem is not as easy as first glance, let's examine a simplified one: 

you are given a sequence A[1], A[2],..., A[N]. On this sequence you have to apply M operations: Add all the elements whose value are in range [l, r] with d or, ask for a query how many element are in range [l, r]. 

 

Input

There are only one test case, Process until the end of the file. The first line of each case contains two numbers, N, M, described as above. And then start from the second line, have N numbers described the sequence's initial value. 

( 1≤ N ≤ 250,000, M ≤ 50,000), |A[i]|≤ 1,000,000,000 .) 

The following M lines described the operation: 

C l r d: Add all the element whose value are in range [l, r] with d. (Redeclare: Not its Position! .. ) Q l r: ask for a query how many elements, whose value are in range [l, r]. 

( l ≤ r, |l|,|r|,|d|≤ 1,000,000,000 ) 

We guarantee every elements are suits 32-integer, and will not cause overflow, even during the running-time. (.. but still be careful ;) Besides, all the test-data are generated randomly. 

 

Output

For each query, print the result. Example

 

Sample Input

10 1010 4 -5 8 8 3 0 -2 4 7C -9 8 2C -4 10 -3C -10 0 5Q -9 -1C -9 -5 8C -7 4 3Q -2 7C -10 -3 2C -4 -1 -6Q 7 10

 

Sample Output

1104
Hint
(In the first example, after the two operations, the sequences are become to, {-4, -3, -2, -1, 0, 11, 12, 13, 14, 15}, so there are no elements whose value are in range [1, 10]. ) 
 

 

又get到了新姿势。。 每隔4000次给排个序重新建树。。 这样做只知道可以优化复杂度

#include <bits/stdc++.h>using namespace std;const int MAXN = 250010;int n,m;int a[MAXN];struct node{    int MAX,MIN; int l,r;    int lazy;}tree[MAXN*4];void push_up(int i){    tree[i].MAX = max(tree[i<<1].MAX, tree[i<<1|1].MAX);    tree[i].MIN = min(tree[i<<1].MIN, tree[i<<1|1].MIN);}void build(int i, int l, int r){    tree[i].l = l; tree[i].r = r;    tree[i].lazy = 0;    if(l == r){        tree[i].MAX = tree[i].MIN = a[l];        return;    }    int mid = (l+r)>>1;    build(i<<1, l, mid); build(i<<1|1, mid+1, r);    push_up(i);}void push_down(int i){    if(tree[i].lazy){        tree[i<<1].MAX += tree[i].lazy;        tree[i<<1].MIN += tree[i].lazy;        tree[i<<1|1].MIN += tree[i].lazy;        tree[i<<1|1].MAX += tree[i].lazy;        tree[i<<1].lazy += tree[i].lazy;        tree[i<<1|1].lazy += tree[i].lazy;        tree[i].lazy = 0;    }}void update(int i, int l, int r, int val){    if(tree[i].MIN > r || tree[i].MAX < l) return;    if(tree[i].MIN >=l && tree[i].MAX <= r){        tree[i].MIN += val;        tree[i].MAX += val;        tree[i].lazy += val;        return;    }    push_down(i);    update(i<<1, l, r, val);    update(i<<1|1, l, r, val);    push_up(i);}int query(int i, int l, int r){    if(tree[i].MIN > r || tree[i].MAX < l) return 0;    if(tree[i].MIN >= l && tree[i].MAX <= r){        return tree[i].r - tree[i].l + 1;     }    push_down(i);    return query(i<<1, l, r) + query(i<<1|1, l, r);}void getA(int i){    if(tree[i].l == tree[i].r){        a[tree[i].l] = tree[i].MAX;        return;    }    push_down(i);    getA(i<<1); getA(i<<1|1);}char cmd[3];int main(){    int x,y,z;    while(cin>>n>>m){        for(int i=1; i<=n; i++) scanf("%d", a+i);        sort(a+1, a+n+1);        build(1, 1, n);        for(int i=1; i<=m; i++){            scanf("%s", cmd);            if(cmd[0] == 'C'){                scanf("%d %d %d", &x, &y, &z);                   update(1, x, y, z);            }            else{                scanf("%d %d", &x, &y);                int res = query(1, x, y);                printf("%d\n", res);            }            if(i % 4000 == 0){                getA(1);                sort(a+1, a+n+1);                build(1, 1, n);            }        }    }    return 0;}