nyoj 15

令res[i][j]表示从str[i]到str[j]需要添加字符的个数；

如果str[k]与str[i]匹配，  res[i][j]=min(res[i][j], res[i+1][k-1]+res[k+1][j]);

如果str[k]与str[j]匹配，  res[i][j]=min(res[i][j],res[i][k-1]+res[k+1][j-1] );

如果str[k]与str[i]和str[j]都不匹配，res[i][j]=min(res[i][j],res[i][k-1]+res[k+1][j]+1 );

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 1234567890
char str[110];
int res[110][110];
int min(int a,int b)
{
    return a < b ? a : b;
}
int check(int i,int j)
{
    if(str[i]=='[' && str[j]==']') return 1;
    if(str[i]=='(' && str[j]==')') return 1;
    return 0;
}
int work(int n)
{
    int i,j,k;
    for(i=1;i<=n;i++) res[i][i]=1;
    for(i=1;i<=n;i++)
    {
        for(j=i-1;j>0;j--)
        {
            res[j][i]=MAX;
            for(k=j;k<=i;k++)
            {
                if(check(j,k)) res[j][i]=min(res[j][i],res[j+1][k-1]+res[k+1][i]);
                if(check(k,i)) res[j][i]=min(res[j][i],res[j][k-1]+res[k+1][i-1]);
                if(!check(k,i) && !check(j,k)) res[j][i]=min(res[j][i],res[j][k-1]+res[k+1][i]+1);
            }
        }
    }
    return res[1][n];
}
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",str+1);
        memset(res,0,sizeof(res));
        printf("%d\n",work(strlen(str+1)));
    }
    return 0;

}