POJ 1141-Brackets Sequence(区间DP)

题目链接：

http://poj.org/problem?id=1141

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32282 Accepted: 9321 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

()[()]

//区间DP#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 1005#define inf 0x3f3f3f3fusing namespace std;char s[maxn];int dp[maxn][maxn];             //dp[i][j]表示区间i~j内需要最少的字符数能够匹配int path[maxn][maxn];           //path[i][j]是表示怎么达到dp[i][j]状态的int peidui(char c1, char c2){if((c1=='(' && c2==')')||(c1=='['&&c2==']'))return 1;return 0;}//输出区间(x,y)的配对结果void print_path(int x, int y){    if(x>y)        return;//只剩下一个,补上完整的if(x==y){if(s[x]=='(' || s[x]==')')printf("()");elseprintf("[]");return;}if(path[x][y]==-1)   //s[x]和s[y]配对{printf("%c", s[x]);print_path(x+1, y-1);printf("%c", s[y]);}else{int k = path[x][y];print_path(x, k);print_path(k+1, y);}}int main(){while(gets(s)){memset(dp, 0, sizeof(dp));int l = strlen(s);for(int i=0; i<l; i++)       //单个字符需要1个dp[i][i] = 1;for(int len=1; len<l; len++)for(int i=0; i<l-len; i++){    int j = i+len;dp[i][j] = inf;if(peidui(s[i], s[j])){if(dp[i+1][j-1]<dp[i][j]){dp[i][j] = dp[i+1][j-1];path[i][j] = -1;}}                for(int k=i; k<j; k++)                {                    if(dp[i][k]+dp[k+1][j] < dp[i][j])                    {                        dp[i][j] = dp[i][k]+dp[k+1][j];                        path[i][j] = k;                    }                }}print_path(0, l-1);printf("\n");}return 0;}