POJ 1141 Brackets Sequence (區間DP)
阿新 • • 發佈:2017-07-25
ive bsp rip mes character har typedef som memset
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.
The input file contains at most 100 brackets (characters ‘(‘, ‘)‘, ‘[‘ and ‘]‘) that are situated on a single line without any other characters among them.
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
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
()[()]
題意:給一串括號序列。依照合法括號的定義,加入若幹括號,使得序列合法。
典型區間DP。設dp[i][j]為從i到j須要加入最少括號的數目。
dp[i][j] = max{ dp[i][k]+dp[k+1][j] } (i<=k<j)
假設s[i] == s[j] , dp[i][j] 還要和dp[i+1][j-1]比較。 枚舉順序依照區間長度枚舉。
由於要求輸出合法序列,就要記錄在原序列在哪些位置進行了添加,設c[i][j]為從i到j的 添加括號的位置,假設不須要添加。那麽c[i][j] 賦為-1,打印時僅僅需遞歸打印就可以。
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; typedef long long LL; const int MAX=0x3f3f3f3f; int n,c[105][105],dp[105][105]; char s[105]; void print(int i,int j) { if( i>j ) return ; if( i == j ) { if(s[i] == '(' || s[i] == ')') printf("()"); else printf("[]"); return ; } if( c[i][j] > 0 ) { // i到j存在添加括號的地方,位置為c[i][j] print(i,c[i][j]); print(c[i][j]+1,j); } else { if( s[i] == '(' ) { printf("("); print(i+1,j-1); printf(")"); } else { printf("["); print(i+1,j-1); printf("]"); } } } void DP() { //區間DP for(int len=2;len<=n;len++) for(int i=1;i<=n-len+1;i++) { int j = i+len-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]; c[i][j] = k; // 記錄斷開的位置 } if( ( s[i] == '(' && s[j] == ')' || s[i] == '[' && s[j] == ']' ) && dp[i][j] > dp[i+1][j-1] ) { dp[i][j] = dp[i+1][j-1]; c[i][j] = -1; //i到j不須要斷開。由於dp[i+1][j-1]的值更小,上面枚舉的k位置都比這個大。所以不再斷開 } } } int main() { scanf("%s",s+1); n = strlen(s+1); memset(c,-1,sizeof(c)); memset(dp,MAX,sizeof(c)); for(int i=1;i<=n;i++) dp[i][i] = 1, dp[i][i-1] = 0; //賦初值 DP(); print(1,n); printf("\n"); return 0; }
POJ 1141 Brackets Sequence (區間DP)