SRM 688 ParenthesesDiv2Hard

给出一段仅由左右括号组成的字符串，和若干个区间(L,R) (各个区间不重叠)，问要使每个区间内的序列都合法，最少需要进行多少次交换（每次交换可以互换任意两个位置上的字符）。

思路

<1>对于每一个的区间一定要满足的是: ( 个数始终大于)于是每次cnt[1]>cnt[0]时就使cnt[1]–,cnt[0]++,res++ 表示把)变成(的个数 res为A；
<2>最终造成 ( 多一些于是需要把(cnt[1]-cnt[0])/2个 ( 换成 ) 为B;
<3>如果A比B大就需要额外的)反之需要(

const int M=10005;class ParenthesesDiv2Hard {public:    bool G[M],use[M];int need[2],sum[2];    int minSwaps(string s, vector <int> L, vector <int> R) {        int n=s.size(),m=L.size(),ans=0;        for(int i=0;i<n;i++)G[i]=(s[i]==')');        for(int i=0;i<m;i++)            for(int j=L[i];j<=R[i];j++)use[j]=1;        for(int i=0;i<n;i++)if(!use[i])sum[G[i]]++;        for(int i=0;i<m;i++){            if((R[i]-L[i]+1)%2)return -1;            int sum=0,res=0;            //res:how many ')' should be change into '('            //sum:the sum of rest '('            for(int j=L[i];j<=R[i];j++){                if(!G[j])sum++;                else sum--;                if(sum<0)sum+=2,res++;            }            sum/=2;            //sum:'(' was rest and should be exchanged sum/2 times            //->sum/2:how many '(' should be change into ')'            ans+=min(sum,res);            if(res>sum)need[0]+=res-sum;            else need[1]+=sum-res;        }        int chang_inside=min(need[0],need[1]);        //min(need[0],need[1]):how many '(',')'can be fixed inside         if(need[0]-chang_inside>sum[0]||need[1]-chang_inside>sum[1])return -1;        ans+=max(need[0],need[1]);        //ans+=min(need[0],need[1])+max(need[0],need[1])-min(need[0],need[1]);        //     which be exchanged inside  ||   the rest '('or ')' also needs one time to fix;        return ans;    }};