hdu 4417 Super Mario/树套树

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=4417

题意很简单，给定一个序列求一个区间 [L, R,]中小于等于H的元素的个数。

好像函数式线段树可解吧，可弱弱的沙茶一直没弄懂其精髓，只好用树套树暴力碾压了

额树套树，线段树的每一个节点套一个sb树。

当查询[l,r]区间中的值小于等于H的个数，先用线段树找到相应的区间，

然后再查询该区间下对应的平衡树中小于等于H的个数，累加即可。

一直以为会超时，结果400+ms就过了，数据应该很弱吧(自己对拍了一组(N,M)10w规模的跑了2s多o(╯□╰)o)。

  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define lc root<<1
  6 #define rc root<<1|1
  7 const int Max_N = 100100;
  8 struct SBT{
  9     int v, s, c;
 10     SBT *ch[2];
 11     inline void set(int _v = 0){
 12         v = _v, c = s = 1;
 13         ch[0] = ch[1] = null;
 14     }
 15     inline void push_up(){
 16         s = ch[0]->s + ch[1]->s + c;
 17     }
 18     inline int cmp(int x) const{
 19         return v == x ? -1 : x > v;
 20     }
 21 }*null, stack[Max_N << 3], *ptr[Max_N << 2];
 22 int sz = 0, sum = 0, arr[Max_N];
 23 void init(){
 24     null = &stack[sz++];
 25     null->v = null->s = null->c = 0;
 26 }
 27 inline void rotate(SBT* &x, int d){
 28     SBT *k = x->ch[!d];
 29     x->ch[!d] = k->ch[d];
 30     k->ch[d] = x;
 31     k->s = x->s;;
 32     x->push_up();
 33     x = k;
 34 }
 35 void Maintain(SBT* &x, int d){
 36     if (x->ch[d] == null) return;
 37     if (x->ch[d]->ch[d]->s > x->ch[!d]->s){
 38         rotate(x, !d);
 39     } else if (x->ch[d]->ch[!d]->s > x->ch[d]->s){
 40         rotate(x->ch[d], d), rotate(x, !d);
 41     } else {
 42         return;
 43     }
 44     Maintain(x, 0), Maintain(x, 1);
 45 }
 46 void insert(SBT* &x, int v){
 47     if (x == null){
 48         x = &stack[sz++];
 49         x->set(v);
 50     } else {
 51         x->s++;
 52         int d = x->cmp(v);
 53         if (-1 == d){
 54             x->c++;
 55             return;
 56         }
 57         insert(x->ch[d], v);
 58         x->push_up();
 59         Maintain(x, d);
 60     }
 61 }
 62 int sbt_rank(SBT *x, int key){
 63     int t, cur;
 64     for (t = cur = 0; x != null;){
 65         t = x->ch[0]->s;
 66         if (key < x->v) x = x->ch[0];
 67         else if (key >= x->v) cur += x->c + t, x = x->ch[1];
 68     }
 69     return cur;
 70 }
 71 void seg_built(int root, int l, int r){
 72     ptr[root] = null;
 73     for (int i = l; i <= r; i++) insert(ptr[root], arr[i]);
 74     if (l == r) return;
 75     int mid = (l + r) >> 1;
 76     seg_built(lc, l, mid);
 77     seg_built(rc, mid + 1, r);
 78 }
 79 void seg_rank(int root, int l, int r, int x, int y, int v){
 80     if (x > r || y < l) return;
 81     if (x <= l && y >= r){
 82         sum += sbt_rank(ptr[root], v);
 83         return;
 84     }
 85     int mid = (l + r) >> 1;
 86     seg_rank(lc, l, mid, x, y, v);
 87     seg_rank(rc, mid + 1, r, x, y, v);
 88 }
 89 int main(){
 90 #ifdef LOCAL
 91     freopen("in.txt", "r", stdin);
 92     freopen("out.txt", "w+", stdout);
 93 #endif
 94     int i, t, n, m, a, b, c, k = 1;
 95     scanf("%d", &t);
 96     while (t--){
 97         sz = 0, init();
 98         scanf("%d %d", &n, &m);
 99         printf("Case %d:\n", k++);
100         for (i = 1; i <= n; i++) scanf("%d", &arr[i]);
101         seg_built(1, 1, n);
102         while (m--){
103             scanf("%d %d %d", &a, &b, &c);
104             sum = 0;
105             seg_rank(1, 1, n, a + 1, b + 1, c);
106             printf("%d\n", sum);
107         }
108     }
109     return 0;
110 }