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 + ;
char str[maxn];
int po[maxn], suffix[maxn], prefix[maxn];
int be[maxn], en[maxn];
int n, q;
int ans[][];
int getAns(int one, int two) {
for (int i = min(en[two] - be[two] + , en[one] - be[one] + ); i >= ; --i) {
if (prefix[be[two] + i - ] == suffix[en[one] - i + ]) {
return i;
}
}
return ;
}
int pos[maxn];
void work() {
scanf("%d%d", &n, &q);
int magic = (int)sqrt(n * 1.0);
int to = ;
for (int i = ; i <= n; ++i) {
scanf("%s", str + en[i - ] + );
int len = strlen(str + en[i - ] + );
be[i] = en[i - ] + , en[i] = en[i - ] + len;
prefix[be[i]] = str[be[i]];
for (int j = be[i] + ; j <= en[i]; ++j) {
prefix[j] = prefix[j - ] * + str[j];
}
suffix[en[i]] = str[en[i]];
for (int j = en[i] - ; j >= be[i]; --j) {
suffix[j] = str[j] * po[en[i] - j] + suffix[j + ];
}
if (len > magic) {
pos[i] = ++to;
}
}
// printf("%s\n", str + 1);
// for (int i = 1; i <= n; ++i) {
// printf("%d %d\n", be[i], en[i]);
// }
for (int i = ; i <= to; ++i) {
for (int j = ; j <= to; ++j) {
ans[i][j] = -;
}
}
while (q--) {
int one, two;
scanf("%d%d", &one, &two);
if (en[one] - be[one] + <= magic || en[two] - be[two] + <= magic) {
printf("%d\n", getAns(one, two));
} else {
if (ans[pos[one]][pos[two]] == -) {
ans[pos[one]][pos[two]] = getAns(one, two);
}
printf("%d\n", ans[pos[one]][pos[two]]);
}
}
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
po[] = ;
for (int i = ; i <= maxn - ; ++i) po[i] = po[i - ] * ;
while (scanf("%d%d", &n, &q) > ) work();
return ;
}

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 + ;
char str[maxn];
LL po[maxn], suffix[maxn], prefix[maxn];
int be[maxn], en[maxn];
int n, q;
int ans[][];
int getAns(int one, int two) {
for (int i = min(en[two] - be[two] + , en[one] - be[one] + ); i >= ; --i) {
if (prefix[be[two] + i - ] == suffix[en[one] - i + ]) {
return i;
}
}
return ;
}
int pos[maxn];
void work() {
int magic = (int)sqrt( * 1.0);
int to = ;
for (int i = ; i <= n; ++i) {
scanf("%s", str + en[i - ] + );
int len = strlen(str + en[i - ] + );
be[i] = en[i - ] + , en[i] = en[i - ] + len;
prefix[be[i]] = str[be[i]];
for (int j = be[i] + ; j <= en[i]; ++j) {
prefix[j] = prefix[j - ] * + str[j];
}
suffix[en[i]] = str[en[i]];
for (int j = en[i] - ; j >= be[i]; --j) {
suffix[j] = str[j] * po[en[i] - j] + suffix[j + ];
}
if (len > magic) {
pos[i] = ++to;
}
}
// printf("%s\n", str + 1);
// for (int i = 1; i <= n; ++i) {
// printf("%d %d\n", be[i], en[i]);
// }
for (int i = ; i <= to; ++i) {
for (int j = ; j <= to; ++j) {
ans[i][j] = -;
}
}
cin >> q;
while (q--) {
int one, two;
scanf("%d%d", &one, &two);
if (en[one] - be[one] + <= magic || en[two] - be[two] + <= magic) {
printf("%d\n", getAns(one, two));
} else {
if (ans[pos[one]][pos[two]] == -) {
ans[pos[one]][pos[two]] = getAns(one, two);
}
printf("%d\n", ans[pos[one]][pos[two]]);
}
}
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
po[] = ;
for (int i = ; i <= maxn - ; ++i) po[i] = po[i - ] * ;
while (scanf("%d", &n) > ) work();
return ;
}
05-23 07:15