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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.
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("\n"); } return 0;}
POJ 1141 输出正确的括号匹配(最少添加)
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