D:Sequence Swapping

最新推荐文章于 2024-09-28 02:27:55 发布

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原文链接：http://www.cnblogs.com/mfys/p/8973477.html

本文介绍了一种解决括号序列问题的动态规划算法，旨在通过交换括号位置获得最大得分。具体步骤包括读取输入序列、计算每一步可能的最大增益，并最终输出可能达到的最大得分。

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 $in order.$

$就是一个dp,dp[i][j]表示第i个左括号扩到第j个时候的最大值。$

$然后更新即可。$

$但是他没到的不能和他已经到的同等对待。$

$玛德智障，$

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e3+88;
char s[N];
int n,p[N],nxt[N];
long long sum[N],v[N];
long long mx[2][N];
int main(){
    int T;
    while(scanf("%d",&T)!=EOF){
    while(T--){
        scanf("%d",&n);
        long long ans=0;
        int tot=0,f=0;
        scanf("%s",s+1);
        for(int i=1;i<=n;++i) scanf("%lld",v+i);
        for(int i=n;i>=1;--i) if(s[i]=='(') p[++tot]=i;
        for(int i=1;i<=n;++i) sum[i]=sum[i-1]+(s[i]==')'?v[i]:0);
        memset(nxt,0,sizeof(nxt)); 
        for(int i=1;i<=n;++i) if(s[i]==')') {
            if(f) nxt[f]=i;
            f=i;
        }
        for(int i=1;i<=n;++i) mx[1][i]=mx[0][i]=0;
        mx[1][0]=mx[0][0]=-(1LL<<60);
        int cur=0;
        for(int i=1;i<=tot;++i) {
            for(int j=p[i]+1;j<=n;++j) if(s[j]==')') mx[cur][j]=v[p[i]]*(sum[j]-sum[p[i]])+mx[cur^1][j];
            for(int j=n;j>=p[i];--j) if(s[j]==')') mx[cur][j]=max(mx[cur][nxt[j]],mx[cur][j]);
            for(int j=p[i]-1;j>=1;--j) if(s[j]==')') mx[cur][j]=max(mx[cur^1][j],mx[cur][nxt[j]]);
            for(int j=1;j<=n;++j) if(s[j]==')') ans=max(ans,mx[cur][j]);
            cur^=1;
        }
        printf("%lld\n",ans);
    }
    }
}