UVA1626-Brackets sequence（动态规划基础）

Problem UVA1626-Brackets sequence

Time Limit: 4500 mSec

Problem Description

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs. The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a single line without any other characters among them.

 Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line. Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

 

 Sample Input

1

([(]

 

Sample Output

()[()]

题解：这个题挺好的，区间dp，可以写成记忆化搜索，容易忽略的地方是如果区间两边的括号匹配，那么可以用中间的部分转移，然后就是普通的分成左区间和右区间进行转移，这个题比较有价值的地方在于打印解的过程，应该学习一下，就是根据结果，逆推回去，这个方便在不用中间记录转移路径，代价就是时间上会有额外的开销，不过一般不至于因此就TLE，因为解一般很少。输入输出有坑，需要用fgets，并且注意fgets会把最后的'\n'读进来，因此真实串的长度需要-1.

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;

 int iCase, dp[maxn][maxn];
 char bra[maxn];
 bool vis[maxn][maxn];

 bool match(char a, char b) {
     if ((a == '(' && b == ')') || (a == '[' && b == ']')) return true;
     return false;
 }

 int DP(int l, int r) {
     if (dp[l][r] >= ) return dp[l][r];
     if (l == r) return dp[l][r] = ;
     if (l > r) return  dp[l][r] = ;

     int &ans = dp[l][r];
     ans = r - l + ;
     if (match(bra[l], bra[r])) {
         ans = min(ans, DP(l + , r - ));
     }
     for (int k = l; k < r; k++) {
         ans = min(ans, DP(l, k) + DP(k + , r));
     }
     return ans;
 }

 void ans_print(int l, int r) {
     if (l > r) return;
     if (l == r) {
         if (bra[l] == '(' || bra[l] == ')') {
             printf("()");
         }
         else {
             printf("[]");
         }
         return;
     }

     int &ans = dp[l][r];
     if (match(bra[l], bra[r]) && ans == dp[l + ][r - ]) {
         printf("%c", bra[l]);
         ans_print(l + , r - );
         printf("%c", bra[r]);
         return;
     }
     else {
         for (int k = l; k < r; k++) {
             if (ans == dp[l][k] + dp[k + ][r]) {
                 ans_print(l, k);
                 ans_print(k + , r);
                 return;
             }
         }
     }
 }

 int main()
 {
     //freopen("input.txt", "r", stdin);
     scanf("%d\n", &iCase);
     while (iCase--) {
         fgets(bra, maxn, stdin);
         memset(dp, -, sizeof(dp));
         int len = strlen(bra);
         int ans = DP(, len - );
         ans_print(, len - );
         printf("\n");
         if (iCase) printf("\n");
         fgets(bra, maxn, stdin);
     }
     return ;
 }