P4396 [AHOI2013]作业

题目链接

luogu4396

思路

唯有水题暖人心
咕了4天，今天跟着std对拍才做出来不得不说题解真的水的一批
先离散化一下
第一问差分询问，权值树状数组套一套就好了 \(nlog_{n}\)
第二问，Emma
莫队，加上树状数组维护修改 \(nlog_{n}sqrt_{n}\)
\(3s\)随便过吧，最慢的也才\(800ms\)

代码

#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int maxn = 1e5 + 7;
int n, m, w[maxn];
int sub[maxn * 3], belong[maxn], lsh[maxn * 3];
struct query {
    int s, t, a, b, id, ans1, ans2, tmp;
} Q[maxn];
bool cmp1(query a, query b) {return a.s < b.s;}
bool cmp2(query a, query b) {return a.t < b.t;}
bool cmp3(query a, query b) {return belong[a.t] == belong[b.t] ? a.s < b.s : a.t < b.t;}
bool cmp4(query a, query b) {return a.id < b.id;}
int read() {
    int x = 0, f = 1; char s = getchar();
    for (; s < '0' || s > '9'; s = getchar()) if (s == '-')f = -1;
    for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}
int sum[maxn*6];
void add(int x, int ad) {
    for (int i = x; i <= 3e5; i += (i&-i)) sum[i] += ad;
}
int query(int x) {
    int ans = 0;
    for (int i = x; i >= 1; i -= (i&-i)) ans += sum[i];
    return ans;
}
inline void add(int x) {
    if (!sub[x]) add(x, 1);
    sub[x]++;
}
inline void delet(int x) {
    sub[x]--;
    if (!sub[x]) add(x, -1);
}
int main() {
    // read
    n = read(), m = read();
    int k = sqrt(n);
    FOR(i, 1, n) lsh[i] = w[i] = read();
    FOR(i, 1, n) belong[i] = (i - 1) / k + 1;
    int js = n;
    FOR(i, 1, m) {
        Q[i].s = read(),Q[i].t = read(),Q[i].a = read(),Q[i].b = read();
        lsh[++js] = Q[i].a,lsh[++js] = Q[i].b, Q[i].id = i;
    }
    sort(lsh + 1, lsh + 1 + n + m * 2);
    int nn = unique(lsh + 1, lsh + 1 + n + 2 * m) - (lsh + 1);
    FOR(i, 1, n) w[i] = lower_bound(lsh + 1, lsh + 1 + nn, w[i]) - lsh;
    FOR(i, 1, m) {
        Q[i].a = lower_bound(lsh + 1, lsh + 1 + nn, Q[i].a) - lsh;
        Q[i].b = lower_bound(lsh + 1, lsh + 1 + nn, Q[i].b) - lsh;
    }
    
    // work1
    js = 1;
    sort(Q + 1, Q + 1 + m, cmp1);
    FOR(i, 1, n) {
        for (; Q[js].s == i; ++js)
            Q[js].tmp = query(Q[js].b) - query(Q[js].a - 1);
        add(w[i], 1);
    }

    js = 1;
    sort(Q + 1, Q + 1 + m, cmp2);
    memset(sum, 0, sizeof(sum));
    FOR(i, 1, n) {
        add(w[i], 1);
        for(;Q[js].t == i;++js)
            Q[js].ans1 = query(Q[js].b) - query(Q[js].a - 1) - Q[js].tmp;
    }
    
    // work2
    sort(Q + 1, Q + 1 + m, cmp3);
    memset(sum, 0, sizeof(sum));
    int l = 1, r = 0;
    FOR(i, 1, m) {
        while (l > Q[i].s) add(w[--l]);
        while (l < Q[i].s) delet(w[l++]);
        while (r < Q[i].t) add(w[++r]);
        while (r > Q[i].t) delet(w[r--]);
        Q[i].ans2 = query(Q[i].b) - query(Q[i].a - 1);
    }

    //print
    sort(Q + 1, Q + 1 + m, cmp4);
    FOR(i, 1, m) printf("%d %d\n", Q[i].ans1, Q[i].ans2);
}