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 = 100 + 10;

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] >= 0) return dp[l][r];
    if (l == r) return dp[l][r] = 1;
    if (l > r) return  dp[l][r] = 0;

    int &ans = dp[l][r];
    ans = r - l + 1;
    if (match(bra[l], bra[r])) {
        ans = min(ans, DP(l + 1, r - 1));
    }
    for (int k = l; k < r; k++) {
        ans = min(ans, DP(l, k) + DP(k + 1, 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 + 1][r - 1]) {
        printf("%c", bra[l]);
        ans_print(l + 1, r - 1);
        printf("%c", bra[r]);
        return;
    }
    else {
        for (int k = l; k < r; k++) {
            if (ans == dp[l][k] + dp[k + 1][r]) {
                ans_print(l, k);
                ans_print(k + 1, r);
                return;
            }
        }
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    scanf("%d\n", &iCase);
    while (iCase--) {
        fgets(bra, maxn, stdin);
        memset(dp, -1, sizeof(dp));
        int len = strlen(bra);
        int ans = DP(0, len - 2);
        ans_print(0, len - 2);
        printf("\n");
        if (iCase) printf("\n");
        fgets(bra, maxn, stdin);
    }
    return 0;
}