topcoder srm 628 div2 250 500

做了一道题，对了，但是还是掉分了。

第二道题也做了，但是没有交上，不知道对错。

后来交上以后发现少判断了一个条件，改过之后就对了。

第一道题爆搜的，有点麻烦了，其实几行代码就行。

250贴代码：

 #include <iostream>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <cstdio>
 #include <algorithm>
 #include <vector>
 #define LL long long
 using namespace std;

 class BishopMove
 {
 public:
     struct node
     {
         int x, y, step;
     } next, pos;
     int howManyMoves(int r1, int c1, int r2, int c2)
     {
         int vis[][], i, a, b;
         memset(vis, , sizeof(vis));
         if(r1==r2 && c1==c2)
             return ;

         queue<node>q;
         vis[r1][c1] = ;
         next.x = r1;
         next.y = c1;
         next.step = ;
         q.push(next);
         while(!q.empty())
         {
             pos = q.front();
             if(pos.x == r2 && pos.y == c2)
                 return pos.step;

             q.pop();
             for(i = ; i <= ; i ++)
             {
                 a = pos.x+i;
                 b = pos.y+i;
                 if(a>=&&a<= && b>= && b<= && vis[a][b]==)
                 {
                     vis[a][b] = ;
                     next.x = a;
                     next.y = b;
                     next.step = pos.step+;
                     q.push(next);
                 }
                 a = pos.x+i;
                 b = pos.y-i;
                 if(a>=&&a<= && b>= && b<= && vis[a][b]==)
                 {
                     vis[a][b] = ;
                     next.x = a;
                     next.y = b;
                     next.step = pos.step+;
                     q.push(next);
                 }
                 a = pos.x-i;
                 b = pos.y+i;
                 if(a>=&&a<= && b>= && b<= && vis[a][b]==)
                 {
                     vis[a][b] = ;
                     next.x = a;
                     next.y = b;
                     next.step = pos.step+;
                     q.push(next);
                 }
                 a = pos.x-i;
                 b = pos.y-i;
                 if(a>=&&a<= && b>= && b<= && vis[a][b]==)
                 {
                     vis[a][b] = ;
                     next.x = a;
                     next.y = b;
                     next.step = pos.step+;
                     q.push(next);
                 }
             }
         }
         return -;
     }
 };

 int main()
 {
     int r1, c1, r2, c2;
     while()
     {
         cin>>r1>>c1>>r2>>c2;
         BishopMove y;
         cout<<y.howManyMoves(r1, c1, r2, c2)<<endl;
     }
     return ;
 }

题意：有三种括号 和 x，x能变成任意的括号，求能否通过变化x使得给的字符串符合括号匹配

500贴代码：

AC代码：

 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <cstdio>
 #include <vector>
 #define LL long long
 using namespace std;

 class BracketExpressions
 {
 public:
    int _max(int c, int d)
    {
        return c > d?c:d;
    }
    int match(char a, char b)
    {
        if(a=='(' && b==')')
        return ;
        if(a=='{' && b=='}')
        return ;
        if(a=='[' && b==']')
        return ;
        if(a=='X' &&(b==']'||b=='}'||b==')'))
        return ;
        if(b=='X' && (a=='['||a=='{'||a=='('))
        return ;
        if(a=='X' && b=='X')
        return ;
        return ;
    }
    string ifPossible(string expression)
    {
        int i, j, k, g, len;
        int d[][];
        string s = expression;
        len = s.size();
        memset(d, , sizeof(d));
        for(i = ; i < len-; i++)
        if(match(s[i], s[i+]))
        d[i][i+] = ;
        for(k = ; k < len; k++)
        {
            for(i = ; i < len-k; i++)
            {
                j = i+k;
                if(match(s[i], s[j])) d[i][j] = d[i+][j-] + ;
                for(g = ; g < k; g++)
                d[i][j] = _max(d[i][i+g]+d[i+g+][j], d[i][j]);
            }
        }
        if(*d[][len-]!=len)
        return "impossible";
        else
        return "possible";
    }
 };