HDU - 4417 Super Mario 主席树+二分

Super Mario

 HDU- 4417Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (thelength is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in L,R    Mario can hit if the maximal height he can jump is H.InputThe first line follows an integer T, the number of test data. For each test data: The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries. Next line contains n integers, the height of each brick, the range is 0,1000000000              .Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)OutputFor each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query. Sample Input

1
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3

Sample Output

Case 1:
4
0
0
3
1
2
0
1
5
1

Source

HDU - 4417

My Solution

题意：给出一个长度为n(1<=n<=1e5)的数组，m(1<=m<=1e5)次询问，每次询问在区间[L,R]中小于等于X的数的个数。

主席树+二分

朴素的主席树是查询区间第k大(从小到大第k大)，所以只需要在每次询问的时候二分的找出在此区间里比X大的最小的数的大小，比如这个值为y(初始化为-1)，则答案为y-1，即把这个比X大的数去掉(因为X可能不止一个，所以要这样处理)，此外如果y未被刷新则说明X是这个区间最大的数了 此时y = R-L+1。

所以时间复杂度 O(n*logn*logn)

空间复杂度 O(n*logn)

应该还有更直接的时间复杂度为O(n*logn)的算法 ^_^

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef long long LL;const int MAXN = 1e5 + 8;//Chairman_Treeint tot, sz;int a[MAXN], b[MAXN], root[20*MAXN], ls[20*MAXN], rs[20*MAXN], sum[20*MAXN];inline void _build(int &o, int l, int r){o = ++ tot;sum[o] = 0;if(l == r) return;int mid = (l + r) >> 1;_build(ls[o], l, mid);_build(rs[o], mid + 1, r);}inline void update(int &o, int l, int r, int last, int p){o = ++ tot;ls[o] = ls[last];rs[o] = rs[last];sum[o] = sum[last] + 1;if(l == r) return;int mid = (l + r) >> 1;if(p <= mid) update(ls[o], l, mid, ls[last], p);else update(rs[o], mid + 1, r, rs[last], p);}inline int _query(int ss, int tt, int l, int r, int k){if(l == r) return l;int mid = (l + r) >> 1;int  cnt = sum[ls[tt]] - sum[ls[ss]];if(k <= cnt) return _query(ls[ss], ls[tt], l, mid, k);else return _query(rs[ss], rs[tt], mid + 1, r, k - cnt);}int main(){#ifdef LOCALfreopen("b.txt", "r", stdin);//freopen("b.out", "w", stdout);#endif // LOCALios::sync_with_stdio(false); cin.tie(0);int T, n, m, i, ql, qr, x, ind, kase = 1;cin >> T;while(T--){cout << "Case " << kase << ":\n"; kase++;cin >> n >> m;for(i = 1; i <= n; i++){cin >> a[i];b[i] = a[i];}sort(b + 1, b + n + 1);sz = unique(b + 1, b + n + 1) - (b + 1);for(i = 1; i <= n; i++){a[i] = lower_bound(b + 1, b + sz + 1, a[i]) - b; //It is based on 1~n..}tot = 0;_build(root[0], 1, sz);for(i = 1; i <= n; i++){update(root[i], 1, sz, root[i-1], a[i]);}int l, r, mid, ansx, ansind;while(m--){cin >> ql >> qr >> x;ql++, qr++;l = 0, r = qr - ql + 1 + 1;ansind = -1;while(l + 1 < r){mid = (l + r) >> 1;ind = _query(root[ql - 1], root[qr], 1, sz, mid);if(b[ind] <= x){l = mid;}else{r = mid;ansind = mid;}}if(ansind == -1) cout << qr - ql + 1 << "\n";else cout << ansind - 1 << "\n";}}return 0;}

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