題意

區間【1， n】上，每個點都有一塊處在某高度的石頭。給 m 個詢問，區間【l， r】上高度不大於 h 的石頭有多少塊。

思路

主席樹模板題

連結

程式碼

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn = 1e5 + 10;

int t, n, m;
int A[maxn];
int node[maxn << 3];
int rot[maxn], lson[maxn << 3
 
], rson[maxn << 3], num[maxn << 3], tt;
int dct[maxn<<1], len;
int a[maxn], b[maxn], h[maxn];

void ins(int l, int r, int pre, int& now, int x)
{
    now = ++tt;

    if(l == r){
        num[now] = num[pre] + 1;
        return;
    }

    lson[now] = lson[pre], rson[now] = rson[pre];

    int
 
 mid = (l + r) >> 1;
    if(x <= mid) ins(l, mid, lson[pre], lson[now], x);
    else ins(mid+1, r, rson[pre], rson[now], x);

    num[now] = num[lson[now]] + num[rson[now]];
}

int query(int l, int r, int a, int b, int x, int y)
{
    if(x > y) return 0;
    if(x <= l && r <= y) return
 
 num[b] - num[a];

    int mid = (l + r) >> 1;

    int res = 0;
    if(x <= mid) res += query(l, mid, lson[a], lson[b], x, y);
    if(y > mid) res += query(mid+1, r, rson[a], rson[b], x, y);

    return res;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    scanf("%d", &t);
    int cas = 0;
    while(t--)
    {
        printf("Case %d:\n", ++cas);
        tt = 0;

        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", A + i);
            dct[i] = A[i];
        }

        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d %d", a+i, b+i, h+i);
            dct[n+i] = h[i];
            a[i]++, b[i]++;
        }

        sort(dct + 1, dct + 1 + n + m);
        len = unique(dct + 1, dct + n + m + 1) - dct - 1;

        for(int i = 1; i <= n; i++)
        {
            int d = lower_bound(dct + 1, dct + len + 1, A[i]) - dct;
            ins(1, len, rot[i-1], rot[i], d);
        }

        for(int i = 1; i <= m; i++)
        {
            int d = lower_bound(dct + 1, dct + len + 1, h[i]) - dct;
            printf("%d\n", query(1, len, rot[a[i]-1], rot[b[i]], 1, d));
        }
    }

    return 0;
}