CSU 1515 Sequence （莫队算法）

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[];
int b[], c[];

int tmp[];
ll so[];

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[];

int update(int x, int d)
{
    int ans = d * ((tmp[x+] + tmp[x-]));
    if (d < ) 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 = ; i <= n; ++i) {
            scanf("%d", &a[i].val);
            a[i].pos = i;
        }
        sort(a+, a++n);
        b[] = ;for (int i = ; i <= n; ++i) {
            if (a[i].val == a[i-].val) b[i] = b[i-];
            else if (a[i].val == a[i-].val + ) b[i] = b[i-]+;
            else b[i] = b[i-] + ;
        }
        for (int i = ; i <= n; ++i) {
            c[ a[i].pos ] = b[i];
        }
        memset(tmp, , sizeof tmp);

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

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