Codeforces 1129D Isolation dp + 分块 （看题解）

Isolation

感觉没有见过这种分块优化的题目啊。。

dp转移很容易就能得出来， 然后分块维护块内信息， 在末尾加入一个数的时候更新块内信息， 更新当前合法的dp值得和。

#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 = 1e5 + 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;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int B = 350;

int n, k, a[N];
int belong[N], bl[N], br[N];
int prePos[N], Map[N];
int c[400][N], lazy[400], g[N];
int dp[N], sumdp;

void changeUp(int L, int R) {
    if(L > R) return;
    if(belong[L] == belong[R]) {
        for(int i = L; i <= R; i++) {
            if(g[i] + lazy[belong[i]] == k) sub(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]++;
            add(c[belong[i]][g[i]], dp[i]);
        }
    } else {
        for(int i = L; i <= br[belong[L]]; i++) {
            if(g[i] + lazy[belong[L]] == k) sub(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]++;
            add(c[belong[i]][g[i]], dp[i]);
        }
        for(int i = bl[belong[R]]; i <= R; i++) {
            if(g[i] + lazy[belong[R]] == k) sub(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]++;
            add(c[belong[i]][g[i]], dp[i]);
        }
        for(int i = belong[L] + 1; i <= belong[R] - 1; i++) {
            sub(sumdp, c[i][k - lazy[i]]);
            lazy[i]++;
        }
    }
}

void changeDn(int L, int R) {
    if(L > R) return;
    if(belong[L] == belong[R]) {
        for(int i = L; i <= R; i++) {
            if(g[i] + lazy[belong[i]] == k + 1) add(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]--;
            add(c[belong[i]][g[i]], dp[i]);
        }
    } else {
        for(int i = L; i <= br[belong[L]]; i++) {
            if(g[i] + lazy[belong[L]] == k + 1) add(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]--;
            add(c[belong[i]][g[i]], dp[i]);
        }
        for(int i = bl[belong[R]]; i <= R; i++) {
            if(g[i] + lazy[belong[R]] == k + 1) add(sumdp, dp[i]);
            sub(c[belong[i]][g[i]], dp[i]);
            g[i]--;
            add(c[belong[i]][g[i]], dp[i]);
        }
        for(int i = belong[L] + 1; i <= belong[R] - 1; i++) {
            add(sumdp, c[i][k + 1 - lazy[i]]);
            lazy[i]--;
        }
    }
}

int main() {
    memset(bl, 0x3f, sizeof(bl));
    memset(br, 0xc0, sizeof(br));
    scanf("%d%d", &n, &k);
    for(int i = 0; i <= n; i++) {
        belong[i] = i / B;
        chkmin(bl[belong[i]], i);
        chkmax(br[belong[i]], i);
    }
    dp[0] = 1;
    c[0][0] = 1;
    sumdp = 1;

    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        prePos[i] = Map[a[i]];
        Map[a[i]] = i;
    }
    for(int i = 1; i <= n; i++) {
        int pos = prePos[i];
        int ppos = prePos[pos];
        changeUp(pos, i - 1);
        changeDn(ppos, pos - 1);
        dp[i] = sumdp;
        add(sumdp, dp[i]);
        add(c[belong[i]][0], dp[i]);
    }
    printf("%d
", dp[n]);
    return 0;
}

/*
*/