[POJ] Brackets Sequence

This problem can be solved elegantly using dynamic programming.

We maintain two arrays:

1. cnt[i][j] --- number of parentheses needed to add within s[i..j] inclusively;
2. pos[i][j] --- position to add the parenthesis within s[i..j] inclusively.

Then there are three cases:

1. cnt[i][i] = 1;
2. If s[i] == s[j], cnt[i][j] = cnt[i + 1][j - 1], pos[i][j] = -1 (no need to add any parenthesis);
3. If s[i] != s[j], cnt[i][j] = min_{k = i, i + 1, ..., j}cnt[i][k] + cnt[k + 1][j], pos[i][j] = k (choose the best position to add the parenthesis).

After computing cnt and pos, we will print the resulting parentheses recursively.

My accepted code is as follows. In fact, I spent a lot timg on debugging the Wrong Answer error due to incorrect input/output. You may try this problem at this link.

 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>

 using namespace std;

 #define INT_MAX 0x7fffffff
 #define vec1d vector<int>
 #define vec2d vector<vec1d >

 void print(char* s, vec2d& pos, int head, int tail) {
     if (head > tail) return;
     if (head == tail) {
         if (s[head] == '(' || s[head] == ')')
             printf("()");
         else printf("[]");
     }
     else if (pos[head][tail] == -) {
         printf("%c", s[head]);
         print(s, pos, head + , tail - );
         printf("%c", s[tail]);
     }
     else {
         print(s, pos, head, pos[head][tail]);
         print(s, pos, pos[head][tail] + , tail);
     }
 }

 void solve(char* s, vec2d& cnt, vec2d& pos) {
     int n = strlen(s);
     for (int i = ; i < n; i++)
         cnt[i][i] = ;
     for (int l = ; l < n; l++) {
         for (int i = ; i < n - l; i++) {
             int j = i + l;
             cnt[i][j] = INT_MAX;
             if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) {
                 cnt[i][j] = cnt[i + ][j - ];
                 pos[i][j] = -;
             }
             for (int k = i; k < j; k++) {
                 if (cnt[i][k] + cnt[k + ][j] < cnt[i][j]) {
                     cnt[i][j] = cnt[i][k] + cnt[k + ][j];
                     pos[i][j] = k;
                 }
             }
         }
     }
     print(s, pos, , n - );
     printf("\n");
 }

 int main(void) {
     char s[];
     while (gets(s)) {
         int n = strlen(s);
         vec2d cnt(n, vec1d(n, ));
         vec2d pos(n, vec1d(n));
         solve(s, cnt, pos);
     }
     return ;
 }