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

题意:

给定M≤N≤105,N−M≤2000,M为原始括号序列s的长度,现要求寻找p,q括号序列
要求|p+s+q|=N,且新序列左右括号数相等,并且任意前缀左括号数大于右括号数
求合法的(p,q)方法数

分析:

考虑dp,括号序列经典的表示方式就是平衡,f[i][j]:=长度i平衡为j的合法括号序列数
转移的时候首先要求括号序列合法(即任意前缀的平衡非负),其次直接根据平衡转移就可以了
预处理出f之后,来累加一下题目要求的(p,q)答案
对于p,要满足p+q括号序列合法(即任意前缀的平衡非负)
假设p的平衡为j,q的平衡为b,最小平衡为minB,显然要满足j+minB≥0
并且由于后面还有q,所以要留够长度,满足j+b≤min(i+m,n−i−m)
剩余的q的平衡应该是−(j+b),由于对称性,乘f[n−i−m][j+b]即可
时间复杂度为O((n−m)2)

代码:

////  Created by TaoSama on 2016-02-22//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m;char s[N];int f[2005][2005];void add(int& x, int y) {    if((x += y) >= MOD) x -= MOD;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d%s", &n, &m, s + 1) == 3) {        f[0][0] = 1;        for(int i = 1; i <= n - m; ++i) {            for(int j = 0; j <= i; ++j) {                if(j - 1 >= 0) add(f[i][j], f[i - 1][j - 1]); //(                if(j + 1 <= i) add(f[i][j], f[i - 1][j + 1]); //(            }        }        int b = 0, minB = INF;        for(int i = 1; i <= m; ++i) {            if(s[i] == '(') ++b;            else --b;            minB = min(minB, b);        }        int ans = 0;        for(int i = 0; i <= n - m; ++i)            for(int j = 0; j <= i; ++j)                if(j + minB >= 0 && j + b <= min(i + m, n - m - i))                    add(ans, 1LL * f[i][j] * f[n - m - i][j + b] % MOD);        printf("%d\n", ans);    }    return 0;}