CSU 1515 Sequence （莫队算法）

Posted 2020-07-21 我不吃饼干呀

题意：给n个数，m个询问。每个询问是一个区间，求区间内差的绝对值为1的数对数。

题解：先离散化，然后莫队算法。莫队是离线算法，先按按询问左端点排序，在按右端点排序。

ps：第一次写莫队，表示挺简单的，不过这题之前乱搞一气一直TLE，莫队还是很强大的。

代码：

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;

struct Node {
    int val;
    int pos;
    bool operator < (const Node x) const {
        return val < x.val;
    }
} a[10005];
int b[10005], c[10005];

int tmp[30000];
ll so[100005];

struct query {
    int l, r, id;
    bool operator < (const query x) const {
        if (l == x.l) return r < x.r;
        return l < x.l;
    }
} q[100005];


int update(int x, int d)
{
    int ans = d * ((tmp[x+1] + tmp[x-1]));
    if (d < 0) tmp[x]--; else tmp[x]++;
    return ans;
}
// 我好菜啊
int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i].val);
            a[i].pos = i;
        }
        sort(a+1, a+1+n);
        b[1] = 1;for (int i = 2; i <= n; ++i) {
            if (a[i].val == a[i-1].val) b[i] = b[i-1];
            else if (a[i].val == a[i-1].val + 1) b[i] = b[i-1]+1;
            else b[i] = b[i-1] + 2;
        }
        for (int i = 1; i <= n; ++i) {
            c[ a[i].pos ] = b[i];
        }
        memset(tmp, 0, sizeof tmp);

        for (int i = 0; i < m; ++i) {
            scanf("%d%d", &q[i].l,&q[i].r);
            q[i].id = i;
        }
        sort(q, q+m);

        int pl = 1, pr = 0;
        ll ans = 0;
        for (int i = 0; i < m; ++i) {
            int id = q[i].id;
            int l = q[i].l;
            int r = q[i].r;
            if (pr < r) for (int i = pr+1; i <= r; ++i) ans += update(c[i], 1);
            else        for (int i = pr; i > r; --i) ans += update(c[i], -1);
            if (pl < l) for (int i = pl; i < l; ++i) ans += update(c[i], -1);
            pr = r, pl = l;
            so[id] = ans;
        }
        for (int i = 0; i < m; ++i)
            printf("%lld\n", so[i]);
    }
    return 0;
}