Super Mario（主席树）

Super Mario 

Mario 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 (the length 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.

Input

The 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.)

Output

For 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

题意：给你N个数，M次询问，每次输入一个区间和一个高度，求这个区间内不超过这个高度的数的个数。
算法：主席树

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 1e5+5;

struct node {
    int l, r, s;
}tree[maxn * 40];

vector<int> v;

int a[maxn];
int cnt;
int root[maxn];

void build(int &cur, int l, int r) {
    cur = ++cnt;
    tree[cur].s = 0;
    if(l == r) {
        return;
    }
    int mid = (l + r) >> 1;
    build(tree[cur].l, l, mid);
    build(tree[cur].r, mid + 1, r);
}

int getId(int x) {
    return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}

void update(int l, int r, int x, int &y, int pos) {
    y = ++cnt;
    tree[y] = tree[x];
    tree[y].s++;
    if(l == r) {
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) {
        update(l, mid, tree[x].l, tree[y].l, pos);
    }  else {
        update(mid + 1, r, tree[x].r, tree[y].r, pos);
    }
}

int query(int l, int r, int x, int y, int h) {
    if(l == r) {
        return tree[y].s - tree[x].s;
    }
    int mid = (l + r) >> 1;
    int ans = 0;    //存储小于该高度的数的数量
    if(h <= mid) {
        ans += query(l, mid, tree[x].l, tree[y].l, h);
    } else {
        ans += tree[tree[y].l].s - tree[tree[x].l].s;    //如果该数在右子树上，那么就需要把左子树上的数都加进去
        ans += query(mid + 1, r, tree[x].r, tree[y].r, h);
    }
    return ans;
}

int main() {
    int T;
    int Case = 1;
    scanf("%d", &T);
    while(T--) {
        int n, m;
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            v.push_back(a[i]);
        }
        sort(v.begin(), v.end());
        v.erase(unique(v.begin(), v.end()), v.end());
        //build(root[0], 1, n);
        int size = v.size();
        for(int i = 1; i <= n; i++) {
            update(1, size, root[i - 1], root[i], getId(a[i]));
        }
        printf("Case %d:\n", Case++);
        while(m--) {
            int left, right, k;
            scanf("%d %d %d", &left, &right, &k);
            left++;        //此处记住需要加1，因为你的主席树是从1开始建立的
            right++;
            int h = upper_bound(v.begin(), v.end(), k) - v.begin();    //找到第一个比k大的数的位置
            int ans = 0;
            if(h > 0) {    
                ans = query(1, size, root[left - 1], root[right], h);
            }
            cout << ans << endl;    
        }
    }
    return 0;
}