CSU 1515 - Sequence (莫队)

Sep 30, 2015


题意

区间l, r内有几对满足

思路

可以发现l+=1, rl, r+-1都可以在O(1)内求出来,就可以用莫队了。

用map搞了好几发,连输入挂都用上了还是TLE。不就多了个log吗(斜眼 后来只能一开始离散化一下。

代码

#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <fstream>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#pragma comment(linker, "/STACK:102400000,102400000")
#include <string>
#include <map>
#include <cmath>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(p, num) memset(p, num, sizeof(p))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid+1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define FOR(i, a, b) for (int i=(a); (i) < (b); (i)++)
#define FOOR(i, a, b) for (int i = (a); (i)<=(b); (i)++)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e4+10;
const int MOD = 1e9+7;
const int dir[][2] = { {0, 1}, {1, 0} };
const int seed = 131;
int cases = 0;
typedef pair<int, int> pii;
 
const int MAXM = 1e5 + 10;
int pos[MAXN];
 
void in(int &x){
    char ch; int minus = 0;
    while (ch=getchar(), (ch<'0'||ch>'9') && ch!='-');
    if (ch == '-') minus = 1, x = 0;
    else x = ch-'0';
    while (ch=getchar(), ch>='0'&&ch<='9') x = x*10+ch-'0';
    if (minus) x = -x;
}
 
struct POINT
{
    int l, r, id;
    bool operator < (const POINT &a) const
    {
        if (pos[l] != pos[a.l]) return l < a.l;
        return r > a.r;
    }
}p[MAXM];
 
LL cur_ans, ans[MAXM], num[MAXN*3];
map<int, int> mp;
int arr[MAXN*3], n, q, tmp[MAXN*3];
 
void update(int pos, int val)
{
    cur_ans += val * num[arr[pos]-1];
    cur_ans += val * num[arr[pos]+1];
    num[arr[pos]] += val;
}
 
void solve()
{
    int cl = 1, cr = 0;
    for (int i = 1; i <= q; i++)
    {
        for (; cr < p[i].r; cr++) update(cr+1, 1);
        for (; cr > p[i].r; cr--) update(cr, -1);
        for (; cl < p[i].l; cl++) update(cl, -1);
        for (; cl > p[i].l; cl--) update(cl-1, 1);
        ans[p[i].id] = cur_ans;
    }
    for (int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
}
 
void handle()
{
    sort(tmp+1, tmp+1+n);
    int len = unique(tmp+1, tmp+1+n)-tmp-1;
    int cnt = 0;
    mp[tmp[1]] = ++cnt;
    for (int i = 2; i <= len; i++)
    {
        if (tmp[i] - tmp[i-1] != 1) ++cnt;
        mp[tmp[i]] = ++cnt;
    }
    for (int i = 1; i <= n; i++) arr[i] = mp[arr[i]];
}
 
int main()
{
    //ROP;
    while (~scanf("%d%d", &n, &q))
    {
        MS(num, 0); mp.clear(); cur_ans = 0;
        int block_size = sqrt(q);
        for (int i = 1; i <= n; i++)
        {
            in(arr[i]);
            tmp[i] = arr[i];
            pos[i] = (i-1) / block_size + 1;
        }
        handle();
        for (int i = 1; i <= q; i++)
        {
            in(p[i].l); in(p[i].r);
            p[i].id = i;
        }
        sort(p+1, p+1+q);
        solve();
    }
    return 0;
}