Codeforces 1175F The Number of Subpermutations

最新推荐文章于 2021-02-24 12:02:33 发布

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最新推荐文章于 2021-02-24 12:02:33 发布

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原文链接：http://www.cnblogs.com/CJLHY/p/10982816.html

本文介绍了一种通过构建特殊的数据结构来高效计算给定数组中子排列数量的算法。使用预处理技术和区间查询优化，该算法能够快速找到满足条件的子序列，特别适用于竞赛编程中的复杂问题。

The Number of Subpermutations

想了想感觉可行解不多，然后就暴力莽就完事了(FST我就删了这个博客逃

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, a[N];
int maxPos[N];
int Map[N];

int Log[N];
struct ST {
    int dp[N][20]; int ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline LL query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
} rmq;

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    rmq.build(n, a, 1);
    int canGo = n;
    for(int i = n; i >= 1; i--) {
        if(Map[a[i]]) chkmin(canGo, Map[a[i]] - 1);
        Map[a[i]] = i;
        maxPos[i] = canGo;
    }
    LL ans = 0;
    for(int i = 1; i <= n; i++) {
        int now = i;
        while(now <= maxPos[i]) {
            if(rmq.query(i, now) == now - i + 1) {
                ans++;
                now++;
            } else {
                now = i + rmq.query(i, now) - 1;
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}

/*
*/