CodeForces - 822E Liar

我们设\(dp[i][j]\)为将串\(s\)前\(i\)个字符分成\(j\)组后能到达串\(t\)的最大位置。

转移方程就是：

\[dp[i][j] = max(dp[i][j], dp[i - 1][j]);\]
\[dp[i + LCP - 1][j + 1] = max(dp[i + LCP - 1][j + 1], dp[i - 1][j] + LCP);\]

其中\(LCP\)就是\(i\)与\(dp[i-1][j]+1\)的\(lcp\)。

我们用后缀数组维护\(lcp\)即可。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 2e5 + 5;

int dp[N][35], x, l1, l2, lg[N], rmq[N][20], n, m, h[N], tax[N], SA[N], tp[N], rk[N], a[N];

int read() {
    int x = 0, f = 1; char S;
    while((S = getchar()) > '9' || S < '0') {
        if(S == '-') f = -1;
        if(S == EOF) exit(0);
    }
    while(S <= '9' && S >= '0') {
        x = (x << 1) + (x << 3) + (S ^ 48);
        S = getchar();
    }
    return x * f;
}

bool cmp(const int x, const int y, const int d) {return tp[x] == tp[y] && tp[x + d] == tp[y + d];}

void Sort() {
    for(int i = 0; i <= m; ++ i) tax[i] = 0;
    for(int i = 1; i <= n; ++ i) ++ tax[rk[tp[i]]];
    for(int i = 1; i <= m; ++ i) tax[i] += tax[i - 1];
    for(int i = n; i >= 1; -- i) SA[tax[rk[tp[i]]] --] = tp[i];
}

void init() {
    char ch[N];
    l1 = read(); scanf("%s", ch);
    for(int i = 1; i <= l1; ++ i) a[i] = ch[i - 1];
    n = l1; a[++ n] = '$';
    l2 = read(); scanf("%s", ch);
    for(int i = n + 1; i <= n + l2; ++ i) a[i] = ch[i - n - 1];
    n += l2;
    m = 122;
    x = read();
}

void Suffix() {
    init();
    for(int i = 1; i <= n; ++ i) rk[i] = a[i], tp[i] = i;
    Sort();
    for(int w = 1, p = 1, i; p < n; m = p, w <<= 1) {
        for(p = 0, i = n - w + 1; i <= n; ++ i) tp[++ p] = i;
        for(i = 1; i <= n; ++ i) if(SA[i] > w) tp[++ p] = SA[i] - w;
        Sort(); swap(rk, tp); rk[SA[1]] = p = 1;
        for(i = 2; i <= n; ++ i) rk[SA[i]] = cmp(SA[i], SA[i - 1], w) ? p : ++ p;
    }
    int j, k = 0;
    for(int i = 1; i <= n; h[rk[i ++]] = k)
        for(k = k ? k - 1 : k, j = SA[rk[i] - 1]; a[i + k] == a[j + k]; ++ k);
    for(int i = 2; i <= n; ++ i) lg[i] = lg[i >> 1] + 1;
    for(int i = 1; i <= n; ++ i) rmq[i][0] = h[i];
    for(int j = 1; (1 << j) <= n; ++ j)
        for(int i = 1; i + (1 << j) - 1 <= n; ++ i)
            rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << j - 1)][j - 1]);
}

int lcp(const int x, const int y) {
    int l = rk[x], r = rk[y];
    if(l == r) return n - SA[l] + 1;
    if(l > r) swap(l, r);
    int t = lg[r - l];
    return min(rmq[l + 1][t], rmq[r - (1 << t) + 1][t]);
}

int main() {
    Suffix();
    for(int i = 1; i <= l1; ++ i)
        for(int j = 0; j < x; ++ j) {
            if(dp[i][j] == l2) {puts("YES"); return 0;}
            dp[i][j] = max(dp[i][j], dp[i - 1][j]);
            int LCP = lcp(i, dp[i - 1][j] + 2 + l1);
            if(LCP > 0) dp[i + LCP - 1][j + 1] = max(dp[i + LCP - 1][j + 1], dp[i - 1][j] + LCP);
            if(dp[i][j] == l2) {puts("YES"); return 0;}
        }
    for(int i = 1; i <= l1; ++ i) if(dp[i][x] == l2) {puts("YES"); return 0;}
    puts("NO");
    return 0;
}

原文地址：https://www.cnblogs.com/AWhiteWall/p/12407526.html