Codeforces Round #343 (Div. 2) C. Famil Door and Brackets

题目链接：

http://codeforces.com/contest/629/problem/C

题意：

长度为n的括号，已经知道的部分的长度为m，现在其前面和后面补充‘(',或')'，使得其长度为n，且每个左括号都能找到右括号与它匹配。

题解：

dp[i][j]表示长度为i，平衡度为j的合法括号序列的总数，这里平衡度定义是‘('比')'多多少个，或')'比’('多多少个。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 2222;
const int maxm = 1e5 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
typedef __int64 LL;

LL dp[maxn][maxn];
void pre() {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int i = 1; i < maxn; i++) {
        dp[i][0] = dp[i - 1][1];
        for (int j = 1; j <= i; j++) {
            dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j + 1]) % mod;
        }
    }
}

int n, m;
char str[maxm];

int main() {
    pre();
    scanf("%d%d", &n, &m);
    scanf("%s", str);
    int l = 0, r = 0, mi = INF;
    for (int i = 0; i < m; i++) {
        if (str[i] == '(') l++;
        else r++;
        mi = min(mi, l - r);
    }
    LL ans = 0;
    for (int i = 0; i <= n - m; i++) {
        for (int j = max(0, -mi); j <= i; j++) {
            int t = j + l - r;
            if (t > n - m - i) continue;
            ans = (ans + dp[i][j] * dp[n - m - i][t] % mod) % mod;
        }
    }
    printf("%I64d
", ans);
    return 0;
}