第15届浙江省赛 D Sequence Swapping(dp)

Sequence Swapping

Time Limit: 1 Second Memory Limit: 65536 KB

BaoBao has just found a strange sequence {< $, >, <, >,, <, >} of length in his pocket. As you can see, each element <, > in the sequence is an ordered pair, where the first element in the pair is the left parenthesis '(' or the right parenthesis ')', and the second element in the pair is an integer.$

As BaoBao is bored, he decides to play with the sequence. At the beginning, BaoBao's score is set to 0. Each time BaoBao can select an integer $, swap the -th element and the -th element in the sequence, and increase his score by, if and only if,'(' and')'.$

BaoBao is allowed to perform the swapping any number of times (including zero times). What's the maximum possible score BaoBao can get?

Input

There are multiple test cases. The first line of the input contains an integer $, indicating the number of test cases. For each test case:$

The first line contains an integer $(), indicating the length of the sequence.$

The second line contains a string $() consisting of '(' and ')'. The -th character in the string indicates, of which the meaning is described above.$

The third line contains $integers (). Their meanings are described above.$

It's guaranteed that the sum of $of all test cases will not exceed .$

Output

For each test case output one line containing one integer, indicating the maximum possible score BaoBao can get.

Sample Input

4
6
)())()
1 3 5 -1 3 2
6
)())()
1 3 5 -100 3 2
3
())
1 -1 -1
3
())
-1 -1 -1

Sample Output

Hint

For the first sample test case, the optimal strategy is to select $in order.$

For the second sample test case, the optimal strategy is to select $in order.$

题解：  
一对括号交换相当于左括号向右移一位，右括号想左移一位，  
所有交换中所有左括号的相对位置不变，右括号同理。  
可以从右边开始枚举每个左括号移动的位置，左括号移动  
到一个点的加分等于此括号到这个点的加分加上之前的括号  
所能到达的点的最大得分。  
代码：  
  
#include<bits/stdc++.h>  
using namespace std;  
#define ll long long  
const int maxn=1003;  
char t[maxn];  
ll a[maxn],dp[maxn][maxn];  
int main()  
{  
    int T;scanf("%d",&T);  
    while(T--)  
    {  
        memset(dp,0,sizeof(dp));  
        int n;scanf("%d",&n);  
        scanf("%s",t+1);  
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);  
        int now=0;  
        ll ma=0;  
        for(int i=n;i>=1;i--)  
        {  
            if(t[i]=='(')  
            {  
                now++;  
                for(int j=i+1;j<=n;j++)  
                {  
                    if(t[j]=='(')dp[now][j]=dp[now][j-1];  
                    else dp[now][j]=a[i]*a[j]+dp[now][j-1];  
                    ma=max(dp[now][j]+dp[now-1][j],ma);  
                }  
                for(int j=i;j<=n;j++)dp[now][j]+=dp[now-1][j];  
                for(int j=n;j>1;j--)dp[now][j-1]=max(dp[now][j-1],dp[now][j]);  
            }  
        }  
        printf("%lld\n",ma);  
    }  
    return 0;  
}