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敲打

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:
", 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
", sum);  
107         }  
108     }  
109     return 0;  
110 }  
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By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明
原文地址:https://www.cnblogs.com/GadyPu/p/4453653.html