BZOJ 4922 Karp-de-Chant Number 动态规划

题目大意：给出一些括号序列，要求选择一些括号序列拼接成一个合法的括号序列，使得总长最大

套路大集合……

首先对于每个括号序列，把左边的左括号和右边的右括号对消，最后能得到一坨这样的东西：
))…))((…((
就是$x$个右括号然后$y$个左括号，记作$(x,y)$

然后考虑假如我们的子集选好了，我们要按照什么顺序拼接才能拼成一个合法的括号序列呢？

BZOJ3709
能拼必须满足当前左括号数$\geq x$，拼完后左括号数$+=y-x$
显然先拼$y\geq x$的，后拼$y </p> <p>先考虑<span class="MathJax_Preview"></span><span class="MathJax" id="MathJax-Element-933-Frame"> <span class="math" id="MathJax-Span-4357" style="width: 3.043em; display: inline-block;"><span style="display: inline-block; position: relative; width: 2.403em; height: 0px; font-size: 125%;"><span style="position: absolute; clip: rect(1.923em 1000em 3.096em -0.477em); top: -2.717em; left: 0.003em;"><span class="mrow" id="MathJax-Span-4358"><span class="mi" id="MathJax-Span-4359" style="font-family: MathJax_Math-italic;">y<span style="display: inline-block; overflow: hidden; height: 1px; width: 0.003em;"></span></span><span class="mo" id="MathJax-Span-4360" style="font-family: MathJax_Main; padding-left: 0.269em;">≥</span><span class="mi" id="MathJax-Span-4361" style="font-family: MathJax_Math-italic; padding-left: 0.269em;">x</span></span><span style="display: inline-block; width: 0px; height: 2.723em;"></span></span></span><span style="border-left: 0.003em solid; display: inline-block; overflow: hidden; width: 0px; height: 1.203em; vertical-align: -0.33em;"></span></span> </span><script type="math/tex" id="MathJax-Element-933">y\geq x$，显然拼了会使左括号数增大，那么能拼就拼，顺序是按照$x$从小到大

然后$y <span class="MathJax" id="MathJax-Element-936-Frame"> <span class="math" id="MathJax-Span-4370" style="width: 1.976em; display: inline-block;"><span style="display: inline-block; position: relative; width: 1.549em; height: 0px; font-size: 125%;"><span style="position: absolute; clip: rect(1.923em 1000em 3.096em -0.371em); top: -2.717em; left: 0.003em;"><span class="mrow" id="MathJax-Span-4371"><span class="mo" id="MathJax-Span-4372" style="font-family: MathJax_Main;">≥</span><span class="mi" id="MathJax-Span-4373" style="font-family: MathJax_Math-italic; padding-left: 0.269em;">y<span style="display: inline-block; overflow: hidden; height: 1px; width: 0.003em;"></span></span></span><span style="display: inline-block; width: 0px; height: 2.723em;"></span></span></span><span style="border-left: 0.003em solid; display: inline-block; overflow: hidden; width: 0px; height: 1.203em; vertical-align: -0.33em;"></span></span> </span><script type="math/tex" id="MathJax-Element-936">\geq y$，拼完后右括号数 $+=x-y$，所以顺序是按照 $y$从小到大，正过来就是从大到小

那么我们把所有的二元组按照上述顺序排序后，以左括号数作为空间跑一遍背包就行了

注意由于物品大小有正有负所以要讨论一下枚举顺序。。。

#include 
#include 
#include 
#include 
#define M 340
using namespace std;

struct abcd
{
    int limit1,limit2,reward,length;
    friend bool operator < (const abcd &x,const abcd &y)
    {
        if(x.reward>=0 && y.reward<0)
            return true;
        if(x.reward<0 && y.reward>=0)
            return false;
        if(x.reward>=0 && y.reward>=0)
            return x.limit1return x.limit2>y.limit2;
    }
}a[M];

int n,sum;
int f[M*M];
int main()
{
    char s[M];
    memset(f,0xef,sizeof f);
    f[0]=0;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        int x=0,y=0;
        scanf("%s",s);
        for(int j=0;s[j];j++)
        {
            if(s[j]=='(')
                y++;
            else
                y?y--:x++;
        }
        a[i].limit1=x;
        a[i].limit2=y;
        a[i].reward=y-x;
        a[i].length=strlen(s);
        sum+=y;
    }
    sort(a+1,a+n+1);

    //for(int i=1;i<=n;i++)
//      cout<

    for(int i=1;i<=n;i++)
    {
        if(a[i].reward<0)
            for(int j=a[i].limit1;j<=sum;j++)
                f[j+a[i].reward]=max(f[j+a[i].reward],f[j]+a[i].length);
        else
            for(int j=sum;j>=a[i].limit1;j--)
                f[j+a[i].reward]=max(f[j+a[i].reward],f[j]+a[i].length);
    }
    cout<0]<return 0;
}