POJ 1141 输出正确的括号匹配(最少添加)

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32174		Accepted: 9291		Special Judge

Description

Let us define a regular brackets sequence in the following way: 

1. Empty sequence is a regular sequence. 
2. If S is a regular sequence, then (S) and [S] are both regular sequences. 
3. If A and B are regular sequences, then AB is a regular sequence. 

For example, all of the following sequences of characters are regular brackets sequences: 

(), [], (()), ([]), ()[], ()[()] 

And all of the following character sequences are not: 

(, [, ), )(, ([)], ([(] 

Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

Input

The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output

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

([(]

Sample Output

()[()]

在上面那个NYoj的基础上添加打印，打印的时候重新检查一下哪个决策最好。好处是节约空间，坏处是打印时代码浮渣，速度稍慢，但是基本上可以忽略不计，因为只有少数状态需要打印。
注意要gets读入，有空串的情况。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000+10;
int dp[maxn][maxn];
char str[maxn];
int n;
bool match(char a,char b) {
    if( a=='('&&b==')'||a=='['&&b==']' )
        return true;
    else
        return false;
    }
void solve() {
    for(int i = 0; i < n; i++) {
        dp[i+1][i] = 0;
        dp[i][i] = 1;
    }

    for(int i = n-2; i >= 0; i--) {
        for(int j = i+1; j < n; j++) {
            dp[i][j] = n;
            if(match(str[i],str[j]))
                dp[i][j] = min(dp[i][j],dp[i+1][j-1]);
            for(int k = i; k < j; k++)
                dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]);

        }
    }
}

void print(int i,int j) {
    if(i>j) return ;
    if(i==j){
       if(str[i]=='('||str[i]==')' )
            printf("()");
       else
        printf("[]");
       return;
    }
    int ans = dp[i][j];
    if(match(str[i],str[j]) && ans==dp[i+1][j-1]) {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
        return;
    }
    for(int k = i; k < j; k++) {
        if(ans==dp[i][k] + dp[k+1][j]) {
            print(i,k);
            print(k+1,j);
            return;
        }
    }
}

int main()
{
//    freopen("in.txt","r",stdin);
//    int T;
//    scanf("%d",&T);
//    getchar();
    while(gets(str)) {
        n = strlen(str);
        solve();
        print(0,n-1);
        printf("
");
    }
    return 0;
}