UVA 1626 Brackets sequence（括号匹配 + 区间DP）

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105116#problem/E

题意：添加最少的括号，让每个括号都能匹配并输出

分析：dp[i][j]表示第i个到第j个需要添加的最少的括号，pos[i][j] = k;表示i到j间第k个需要加括号；

如果str[i]和str[j]匹配，那么dp[i][j] = max(dp[i + 1][j - 1], dp[i][j]);

如果str[i]和str[j]不匹配，那么dp[i][j] = max(dp[i][j], dp[i][k]+dp[k + 1][j]);     i<= k < j;之前想过k为什么不能等于j,

 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <cstdio>
 using namespace std;
 const int MAX = ;
 const int INF = 0x3f3f3f3f;
 int dp[MAX][MAX],pos[MAX][MAX];
 char str[MAX];
 int n;
 void DP()
 {
     n = strlen(str);

     memset(dp, , sizeof(dp));
     for(int i = ; i < n; i++)
         dp[i][i] = ;
     for(int p = ; p < n; p++)
     {
         for(int i = ; i < n; i++)
         {
             int j = i + p;
             if(j >= n)
                 break;
             dp[i][j] = INF;
             if( (str[i] == '(' && str[j] == ')') || ( str[i] == '[' && str[j] == ']' ) )
             {
                 if(dp[i][j] > dp[i + ][j - ])
                     dp[i][j] = dp[i + ][j - ];
             }
             pos[i][j] = -;
             for(int k = i; k < j; k++)
             {
                 int temp = dp[i][k] + dp[k + ][j];
                 if(temp < dp[i][j])
                 {
                     dp[i][j] = temp;
                     pos[i][j] = k;
                 }
             }
         }
     }
 }
 void Print(int beg, int End)
 {
     if(beg > End)
         return;
     if(beg == End)
     {
         if(str[beg] == '(' || str[beg] == ')')
             printf("()");
         if(str[beg] == '[' || str[beg] == ']')
             printf("[]");
     }
     else
     {
         if(pos[beg][End] == -)
         {
             printf("%c", str[beg]);
             Print(beg + , End - );
             printf("%c", str[End]);
         }
         else
         {
             Print(beg, pos[beg][End]);
             Print(pos[beg][End] + , End);
         }
     }
 }
 int main()
 {
     int t;
     scanf("%d", &t);
     getchar();
     getchar();
     for(int i = ; i < t; i++)
     {
         gets(str);
         DP();
         Print(, n - );
         printf("\n");
         if(i < t - )
             printf("\n");
         getchar();
     }

     return ;
 }