POJ 1141 Brackets Sequence (线性dp 括号匹配 经典题)

#include <cstdio>
#include <cstring>
int const INF = 0xfffffff;
int const MAX = 105;
int dp[MAX][MAX], path[MAX][MAX];
char s[MAX];

void Print(int i, int j)
{
    if(i > j) //无效位置
        return;
    if(i == j)  //遇单个字符输出匹配后的结果
    {
        if(s[i] == '(' || s[i] == ')')
            printf("()");
        else
            printf("[]");
    }
    else if(path[i][j] == -1) //若i到j已经匹配，输出左边，递归中间再输出右边
    {
        printf("%c", s[i]);
        Print(i + 1, j - 1);
        printf("%c", s[j]);
    }
    else  //否则，递归输出分割点两边
    {
        Print(i, path[i][j]);
        Print(path[i][j] + 1, j);
    }
}

int main()
{
    while(gets(s))
    {
        int n = strlen(s);
        if(n == 0)
        {
            printf("\n");
            continue;
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++)
            dp[i][i] = 1;
        for(int l = 1; l < n; l++)
        {
            for(int i = 0; i < n - l; i++)
            {
                int j = i + l;
                dp[i][j] = INF;
                if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))
                {
                    dp[i][j] = dp[i + 1][j - 1];
                    path[i][j] = -1;
                }
                for(int k = i; k < j; k++)
                {
                    if(dp[i][j] > dp[i][k] + dp[k + 1][j])
                    {
                        dp[i][j] = dp[i][k] + dp[k + 1][j];
                        path[i][j] = k;
                    }
                }
            }
        }
        Print(0, n - 1);
        printf("\n");
    }
}