Codeforces 822E Liar dp + SA (看题解)

刚开始感觉只要开个dp[ i ][ j ][ 0 / 1 ]表示处理了s的前 i 个用了 k 段， i 是否是最后一段的最后一个字符的 t串最长匹配长度，

然后wa24, 就gg了。感觉这个转移感觉很对，但是实际上不对。。。比如s = ababcde, t = abcde, x = 1, 转移会出现问题。

我们可以用dp[ i ][ j ]表示处理了 s 串的前 i 个，用了 j 段的最大匹配长度，我们转移的时候时候肯定是在后面接lcp，套个sa就好啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#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 ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T> bool chkmax(T& a, T b) {
    return a < b ? a = b, true : false;
}
template<class T> bool chkmin(T& a, T b) {
    return a > b ? a = b, true : false;
}

int Log[N];
struct ST {
    int dp[N][20], 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]);
    }
    int query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y - (1 << k) + 1][k]);
    }
};

int r[N], sa[N], _t[N], _t2[N], c[N], rk[N], lcp[N];
void buildSa(int *r, int n, int m) {
    int i, j = 0, k = 0, *x = _t, *y = _t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = r[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++) {
            if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
                x[sa[i]] = p - 1;
            else x[sa[i]] = p++;
        }
        if(p >= n) break;
        m = p;
     }
     for(i = 1; i < n; i++) rk[sa[i]] = i;
     for(i = 0; i < n - 1; i++) {
        if(k) k--;
        j = sa[rk[i] - 1];
        while(r[i + k] == r[j + k]) k++;
        lcp[rk[i]] = k;
     }
}


int n, m, x, tot, mxc = 256;
int dp[N][32], B;
char s[N], t[N];
ST rmq;

int getLcp(int i, int j) {
    i = rk[i];
    j = rk[j + n + 1];
    if(i > j) swap(i, j);
    return rmq.query(i + 1, j);
}

int main() {
    scanf("%d%s", &n, s);
    scanf("%d%s", &m, t);
    scanf("%d", &x);
    for(int i = 0; s[i]; i++) r[tot++] = s[i];
    r[tot++] = mxc++;
    for(int i = 0; t[i]; i++) r[tot++] = t[i];
    r[tot] = 0;
    buildSa(r, tot + 1, mxc);
    rmq.build(tot, lcp, -1);
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    int LCP = getLcp(0, 0);
    if(LCP) dp[LCP - 1][1] = LCP;
    for(int i = 0; i < n - 1; i++) {
        for(int j = 0; j <= x; j++) {
            if(~dp[i][j]) {
                chkmax(dp[i + 1][j], dp[i][j]);
                if(j == x) continue;
                int LCP = getLcp(i + 1, dp[i][j]);
                if(LCP) chkmax(dp[i + LCP][j + 1], dp[i][j] + LCP);
            }
        }
    }
    bool flag = false;
    for(int i = 0; i < n; i++)
        for(int j = 0; j <= x; j++)
            if(dp[i][j] == m) flag = true;
    puts(flag ? "YES" : "NO");
    return 0;
}

/*
*/