字符串模拟赛T2

// source code from laekov for c0x17
#define PRID "fkqh"
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
const int maxn = 200009;
 
int n, l[maxn], q[maxn], vx[maxn], vy[maxn];
char a[maxn];
 
void manacher() {
    l[0] = 1;
    for (int i = 1, j = 0; i < (n << 1) - 1; ++ i) {
        int r = ((j + 1) >> 1) + l[j] - 1;
        int p = i >> 1, q = i - p;
        l[i] = (r >= q) ? min(r - q + 1, l[(j << 1) - i]) : 0;
        while (p - l[i] >= 0 && q + l[i] < n && a[p - l[i]] == a[q + l[i]])
            ++ l[i];
        if (q + l[i] - 1 > r)
            j = i;
    }
}
 
#define getLeft(x) (((x)>>1)-l[x]+1)
#define getRight(x) ((((x)+1)>>1)+l[x]-1)
 
void dp(bool d) {
    static int q[maxn];
    int hd = 0, tl = 0;
    if (!d) {
        for (int i = 0; i < n; ++ i) {
            if (!tl || getRight(i << 1) > getRight(q[tl - 1]))
                q[tl ++] = (i << 1);
            while (getRight(q[hd]) < i)
                ++ hd;
            vx[i] = (i << 1) - q[hd] + 1;
            if (i < n - 1 && getRight((i << 1) + 1) > getRight(q[tl - 1]))
                q[tl ++] = (i << 1) + 1;
        }
    }
    else {
        for (int i = n - 1; i >= 0; -- i) {
            if (!tl || getLeft(i << 1) < getLeft(q[tl - 1]))
                q[tl ++] = (i << 1);
            while (getLeft(q[hd]) > i)
                ++ hd;
            vy[i] = q[hd] - (i << 1) + 1;
            if (i && getLeft((i << 1) -1) < getLeft(q[tl - 1]))
                    q[tl ++] = (i << 1) - 1;
        }
    }
}
 
int main(int argc, char* args[]) {
    if (argc < 2 || strcmp(args[1], "-nf")) {
        freopen(PRID ".in", "r", stdin);
        freopen(PRID ".out", "w", stdout);
    }
    scanf("%s", a);
    n = strlen(a);
    manacher();
    dp(0);
    dp(1);
    int ans = 2;
    for (int i = 1; i < n; ++ i)
        ans = max(ans, vx[i - 1] + vy[i]);
    printf("%d
", ans);
}
原文地址:https://www.cnblogs.com/hyfer/p/5689475.html