HDU 4417 Super Mario（主席树 区间不超过k的个数）题解

题意：问区间内不超过k的个数

思路：显然主席树，把所有的值离散化一下，然后主席树求一下小于等于k有几个就行。注意，他给你的k不一定包含在数组里，所以问题中的询问一起离散化。

代码：

#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 10;
const int M = maxn * 30;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
int a[maxn], root[maxn], tot;
int n, m;
vector<int> vv;
int getId(int x){
    return lower_bound(vv.begin(), vv.end(),x) - vv.begin() + 1;
}
struct node{
    int lson, rson;
    int sum;
}T[maxn * 20];
void init(){
    memset(T, 0, sizeof(T));
    tot = 0;
    vv.clear();
}
int lowbit(int x){
    return x&(-x);
}
void update(int l, int r, int &now, int pre, int v, int pos){
    T[++tot] = T[pre], T[tot].sum += v, now = tot;
    if(l == r) return;
    int m = (l + r) >> 1;
    if(m >= pos)
        update(l, m, T[now].lson, T[pre].lson, v, pos);
    else
        update(m + 1, r, T[now].rson, T[pre].rson, v, pos);
}
int query(int l, int r, int now, int pre, int v){
    if(l == r){
        if(v >= l)
            return T[now].sum - T[pre].sum;
        return 0;
    }
    if(r <= v)
        return T[now].sum - T[pre].sum;
    int m = (l + r) >> 1;
    int sum = 0;
    if(v < m){
        return query(l, m, T[now].lson, T[pre].lson, v);
    }
    else{
        sum = query(m + 1, r, T[now].rson, T[pre].rson, v);
        return T[T[now].lson].sum - T[T[pre].lson].sum + sum;
    }
}
struct ask{
    int l, r, k;
}q[maxn];
int main(){
    int T, ca = 1;
    scanf("%d", &T);
    while(T--){
        init();
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            vv.push_back(a[i]);
        }
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].k);
            q[i].l++, q[i].r++;
            vv.push_back(q[i].k);
        }
        sort(vv.begin(), vv.end());
        vv.erase(unique(vv.begin(), vv.end()), vv.end());
        for(int i = 1; i <= n; i++){
            update(1, vv.size(), root[i], root[i - 1], 1, getId(a[i]));
        }
        printf("Case %d:\n", ca++);
        for(int i = 1; i <= m; i++){
            printf("%d\n", query(1, vv.size(), root[q[i].r], root[q[i].l - 1], getId(q[i].k)));
        }
    }
    return 0;
}