Harry And Biological Teacher 分块 + 字符串hash

http://acm.hdu.edu.cn/showproblem.php?pid=5069

首先判断suffix和prefix最长多少可以直接暴力枚举长度然后 +　ｈａｓｈ可以立马判断是否相等，复杂度O(lenstr)

现在知道总长度 <= 1e5，  magic = sqrt(lenstr)

那么，

如果长度是 <= magic的，最多1e5个询问，每次询问我暴力也是才1e5 * sqrt(1e5)的复杂度。

如果长度是 >= magic的，

第一个极端情况是有magic个不同的串，这样就可以有magic^2个不同的询问，但是这样每个串的长度最长是magic，这样暴力也是才magic * magic * magic  = 1e5 * sqrt(1e5)的复杂度。

第二个极端情况是只有2个串，每个串长度都很长，1e5 / 2 的长度，这样暴力一次就需要O(1e5)的复杂度了。但是如果这样的话，不同的询问就会很少，可以用个ans数组存起来,ans[a][b]，这样总复杂度也是 <= 1e5 * sqrt(1e5)的复杂度。

所以总复杂度是nsqrtn

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 1e5 + 20;
char str[maxn];
int po[maxn], suffix[maxn], prefix[maxn];
int be[maxn], en[maxn];
int n, q;
int ans[444][444];
int getAns(int one, int two) {
    for (int i = min(en[two] - be[two] + 1, en[one] - be[one] + 1); i >= 1; --i) {
        if (prefix[be[two] + i - 1] == suffix[en[one] - i + 1]) {
            return i;
        }
    }
    return 0;
}
int pos[maxn];
void work() {
    scanf("%d%d", &n, &q);
    int magic = (int)sqrt(n * 1.0);
    int to = 0;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", str + en[i - 1] + 1);
        int len = strlen(str + en[i - 1] + 1);
        be[i] = en[i - 1] + 1, en[i] = en[i - 1] + len;
        prefix[be[i]] = str[be[i]];
        for (int j = be[i] + 1; j <= en[i]; ++j) {
            prefix[j] = prefix[j - 1] * 131 + str[j];
        }
        suffix[en[i]] = str[en[i]];
        for (int j = en[i] - 1; j >= be[i]; --j) {
            suffix[j] = str[j] * po[en[i] - j] + suffix[j + 1];
        }
        if (len > magic) {
            pos[i] = ++to;
        }
    }
//    printf("%s
", str + 1);
//    for (int i = 1; i <= n; ++i) {
//        printf("%d %d
", be[i], en[i]);
//    }
    for (int i = 1; i <= to; ++i) {
        for (int j = 1; j <= to; ++j) {
            ans[i][j] = -1;
        }
    }
    while (q--) {
        int one, two;
        scanf("%d%d", &one, &two);
        if (en[one] - be[one] + 1 <= magic || en[two] - be[two] + 1 <= magic) {
            printf("%d
", getAns(one, two));
        } else {
            if (ans[pos[one]][pos[two]] == -1) {
                ans[pos[one]][pos[two]] = getAns(one, two);
            }
            printf("%d
", ans[pos[one]][pos[two]]);
        }
    }
}
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    po[0] = 1;
    for (int i = 1; i <= maxn - 20; ++i) po[i] = po[i - 1] * 131;
    while (scanf("%d%d", &n, &q) > 0) work();
    return 0;
}

https://www.nowcoder.com/acm/contest/94/E

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef unsigned long long int LL;
const int maxn = 1e6 + 20;
char str[maxn];
LL po[maxn], suffix[maxn], prefix[maxn];
int be[maxn], en[maxn];
int n, q;
int ans[1000][1000];
int getAns(int one, int two) {
    for (int i = min(en[two] - be[two] + 1, en[one] - be[one] + 1); i >= 1; --i) {
        if (prefix[be[two] + i - 1] == suffix[en[one] - i + 1]) {
            return i;
        }
    }
    return 0;
}
int pos[maxn];
void work() {
    int magic = (int)sqrt(300000 * 1.0);
    int to = 0;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", str + en[i - 1] + 1);
        int len = strlen(str + en[i - 1] + 1);
        be[i] = en[i - 1] + 1, en[i] = en[i - 1] + len;
        prefix[be[i]] = str[be[i]];
        for (int j = be[i] + 1; j <= en[i]; ++j) {
            prefix[j] = prefix[j - 1] * 131 + str[j];
        }
        suffix[en[i]] = str[en[i]];
        for (int j = en[i] - 1; j >= be[i]; --j) {
            suffix[j] = str[j] * po[en[i] - j] + suffix[j + 1];
        }
        if (len > magic) {
            pos[i] = ++to;
        }
    }
//    printf("%s
", str + 1);
//    for (int i = 1; i <= n; ++i) {
//        printf("%d %d
", be[i], en[i]);
//    }
    for (int i = 1; i <= to; ++i) {
        for (int j = 1; j <= to; ++j) {
            ans[i][j] = -1;
        }
    }
    cin >> q;
    while (q--) {
        int one, two;
        scanf("%d%d", &one, &two);
        if (en[one] - be[one] + 1 <= magic || en[two] - be[two] + 1 <= magic) {
            printf("%d
", getAns(one, two));
        } else {
            if (ans[pos[one]][pos[two]] == -1) {
                ans[pos[one]][pos[two]] = getAns(one, two);
            }
            printf("%d
", ans[pos[one]][pos[two]]);
        }
    }
}
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    po[0] = 1;
    for (int i = 1; i <= maxn - 20; ++i) po[i] = po[i - 1] * 131;
    while (scanf("%d", &n) > 0) work();
    return 0;
}