poj 1141 Brackets Sequence (区间DP)
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Brackets Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 25893 | Accepted: 7295 | 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.
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题目
#includpe<iostream> #includpe<stdpio.h> #includpe<string.h> using namespace stdp; #dpefine N 105 #dpefine inf 1e9 char s[N]; int dp[N][N],pos[N][N]; void work(int len) { int i,j,k; char left,right; for(i=0;i<len;i++) //若仅仅有一个字符则必须加入一个和它匹配 dp[i][i]=1; for(k=1;k<len;k++) //枚举间隔长度从1到Len-1 { for(i=0;i<len-k;i++) //从起始位置開始 { dp[i][i+k]=inf; left=s[i]; right=s[i+k]; if(left==‘(‘&&right==‘)‘||left==‘[‘&&right==‘]‘) { dp[i][i+k]=dp[i+1][i+k-1]; pos[i][i+k]=-1; } for(j=i;j<i+k;j++) { if(dp[i][j]+dp[j+1][k+i]<dp[i][i+k]) { dp[i][i+k]=dp[i][j]+dp[j+1][i+k]; pos[i][i+k]=j; } } } } } void show(int i,int j) { if(i>j) return ; if(i==j) { if(s[i]==‘(‘||s[i]==‘)‘) printf("()"); else printf("[]"); } else { if(pos[i][j]==-1) { printf("%c",s[i]); show(i+1,j-1); printf("%c",s[j]); } else { show(i,pos[i][j]); show(pos[i][j]+1,j); } } } int main() { while(gets(s)!=NULL) //不能用scanf { int len=strlen(s); work(len); show(0,len-1); puts(""); } return 0; }
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