CF1015F Bracket Substring(动态规划)
CF1015F Bracket Substring
这题一眼GT考试
题意: 求有多少种合法的括号序列要求字符串S是其字串
数据范围大概是一个\(\Theta(n^3)\)的算法
从前往后填括号, 设\(F[i][j][k]\)表示填到第\(i\)位, "和"为\(j\), 匹配了\(S k\)位, 其中和的定义为将左括号看做1, 右括号看作-1, 一个合法的括号序列的前缀和总大于等于零且本身的和为零(一个左括号一定匹配一个右括号)
\(dp\)时只需枚举下一位填\('('\)或\(')'\)即可, 但要注意的是匹配\(S\)串时从第\(k\)位不匹配不是直接重新从0开始匹配而是求出它的\(kmp\)数组, 找到上一个\(border\), 可以参考代码
#pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
template <typename T>
void read(T &x) {
x = 0; bool f = 0;
char c = getchar();
for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
for (;isdigit(c);c=getchar()) x=x*10+(c^48);
if (f) x=-x;
}
template <typename T>
void write(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar('0' + x % 10);
}
const int N = 305;
char s[N];
ll n, m;
int g[N][2], nxt[N];
void KMP(void) {
int j = nxt[1] = 0;
for (int i = 2;i <= m; i++) {
while (j && s[i] != s[j+1]) j = nxt[j];
if (s[i] == s[j+1]) j++;
nxt[i] = j;
}
for (int i = 0;i < m; i++) {
int k = i, p = i;
while (k && s[k+1] != '(') k = nxt[k];
while (p && s[p+1] != ')') p = nxt[p];
if (s[k+1] == '(') k++;
if (s[p+1] == ')') p++;
g[i][1] = k, g[i][0] = p;
}
g[m][0] = g[m][1] = m;
}
const int P = 1e9+7;
ll f[N][N][N];
inline void add(ll &x, ll y) {
x += y; if (x >= P) x -= P;
}
int main() {
read(n); int al = n << 1;
scanf ("%s", s + 1);
m = strlen(s + 1);
KMP();
f[0][0][0] = 1;
for (int i = 0; i < al; i++) {
for (int j = 0; j <= n; j++) {
for (int k = 0; k <= m; k++) {
ll t = f[i][j][k];
if (j) add(f[i+1][j-1][g[k][0]], t);
add(f[i+1][j+1][g[k][1]], t);
}
}
}
cout << f[al][0][m] << endl;
return 0;
}