UVA 2451 Brackets sequence

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=98&page=show_problem&problem=452

用dp[i][j] 记录一段序列，starts from index i, ends with index j，需要添加的char的个数。对于一段序列i~j，如果i, j 配对的话，那么dp[i][j]=dp[i+1][j-1].如果i, j不配对的话，那就需要对序列进行分割，dp[i][j]=dp[i][k]+dp[k+1][j]，对k进行for loop找到最佳的k。但是当i, j配对时，并不代表dp[i+1][j-1]是最佳的结果，仍旧需要寻找最佳的k。代码如下：

 #include <iostream>
 #include <math.h>
 #include <stdio.h>
 #include <cstdio>
 #include <algorithm>
 #include <string.h>
 #include <string>
 #include <sstream>
 #include <cstring>
 #include <queue>
 #include <vector>
 #include <functional>
 #include <cmath>
 #include <set>
 #define SCF(a) scanf("%d", &a)
 #define IN(a) cin>>a
 #define FOR(i, a, b) for(int i=a;i<b;i++)
 #define Infinity 9999999
 typedef long long Int;
 using namespace std;

 int path[][];
 char brackets[];
 void sprint(int s, int e)
 {
     if (s > e)
         return;
     if (s == e)
     {
         if (brackets[s] == '(' || brackets[s] == ')')
             printf("()");
         else
             printf("[]");
     }
     else
     {
         if (path[s][e] == -)
         {
             printf("%c", brackets[s]);
             sprint(s + , e - );
             printf("%c", brackets[e]);
         }
         else
         {
             sprint(s, path[s][e]);
             sprint(path[s][e] + , e);
         }
     }
 }

 int main()
 {
     int N;
     SCF(N);
     cin.get();
     int len = ;
     int dp[][];
     FOR(casenum, , N)
     {
         cin.get();
         cin.getline(brackets, );
         len = ;
         for (len = ; brackets[len] != '\0'; len++);

         if (casenum)
             printf("\n");
         if (len == )
         {
             printf("\n");
             continue;
         }
         FOR(i, , len)
         {
             FOR(j, , len)
                 dp[i][j] = ;
         }
         FOR(i, , len)
             dp[i][i] = ;
         int end = ;

         FOR(clen, , len)
         {
             FOR(start, , len - clen)
             {
                 end = start + clen;
                 dp[start][end] = Infinity;

                 if ((brackets[start] == '(' && brackets[end] == ')') || (brackets[start] == '[' && brackets[end] == ']'))
                 {
                     dp[start][end] = dp[start + ][end - ];
                     path[start][end] = -;
                 }

                 for (int k = start; k < end; k++)
                 {
                     if (dp[start][end] > dp[start][k] + dp[k + ][end])
                     {
                         dp[start][end] = dp[start][k] + dp[k + ][end];
                         path[start][end] = k;
                     }
                 }

             }
         }
         sprint(, len - );
         printf("\n");
     }

     return ;
 }