ZOJ-4027：Sequence Swapping（DP）

Sequence Swapping

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Time Limit: 1 Second       Memory Limit: 65536 KB

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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

24
21
0
2

Hint

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

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

思路：可以想到，对与某个位置为x的'('，能与其交换的')'取决于所有位置为y的'('的交换情况(y>x)。也就是只有后面的'('交换过来的')'，你才能去交换，不然中间隔着一个'('，是无法交换的。那么以d[i][j]表示从i开始，把[i,j]里的')'都与i交换过来的最大值。

当s[i]=')'，d[i][j]=d[i+1][j]。

当s[i]='('，d[i][j]=max(d[i][j+1],d[i+1][j]+(sum[j]-sum[i])*v[i])。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e3+10;
const long long MOD=1e9;
const double PI=acos(-1.0);
typedef long long ll;
char s[MAX];
int v[MAX];
ll d[MAX][MAX],sum[MAX];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        scanf("%s",s+1);
        for(int i=1;i<=n;i++)scanf("%d",&v[i]);
        for(int i=1;i<=n;i++)sum[i]=sum[i-1]+(s[i]==')')*v[i];
        ll ans=0;
        memset(d,0,sizeof d);
        for(int i=n;i>=1;i--)
        {
            d[i][n+1]=-1e16;
            if(s[i]==')')
            {
                for(int j=n;j>=1;j--)d[i][j]=d[i+1][j];
                continue;
            }
            for(int j=n;j>=i;j--)d[i][j]=max(d[i][j+1],d[i+1][j]+(sum[j]-sum[i])*v[i]);
            for(int j=i-1;j>=1;j--)d[i][j]=max(d[i][j+1],d[i+1][j]);         //当j<i时，无法交换
            for(int j=n;j>=1;j--)ans=max(ans,d[i][j]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}