POJ 1141 Brackets Sequence (interval dp)

Language:DefaultBrackets Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 26424 Accepted: 7451 Special Judge Description Let us define a regular brackets sequence in the following by: 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 is regular brackets sequences: (), [], (()), ([]), ()[], ()[()] And all of the following character sequences is not: (, [, ), ) (, ([)], ([(] Some sequence of characters ' (', ') ', ' [', and '] ' is given. You is to find the shortest possible regular brackets sequence, that contains the given character sequence as a Subsequen Ce. Here, a string A1 A2 ... is called a subsequence of the string B1 b2 ... bm, if there exist such indices 1 = I1 < I2 < ... < in = m, which AJ = bij for all 1 = j = N. Input The input file contains at most of brackets (characters ' (', ') ', ' [' and '] ') that is situated on a single line without Any other characters among them. Output Write to the output file A contains some regular brackets sequence this has the minimal possible length a nd contains the given sequence as a subsequence. Sample Input ([(] Sample Output ()[()] Source Northeastern Europe 2001

Interval DP

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace STD; #define N 305#define INF 0x3f3f3f3fint Dp[n][n],vis[n][n];char a[n];inline bool Judge (int I,int j) {if (a[i]== ' (' & &a[j]== ') ' | | a[i]== ' [' &&a[j]== '] ') return True;return false;} void print (int i,int j) {if (i>j) return; if (i==j) {if (a[i]== ' | | | a[i]== ') printf ("()"); elseprintf ("[]"); return;}    if (vis[i][j]==-1) {printf ("%c", A[i]);p rint (i+1,j-1);p rintf ("%c", A[j]); return;}    Print (I,vis[i][j]); Print (VIS[I][J]+1,J);} int main () {int i,j;while (gets (a)) {int Len=strlen (a); if (len==0) {printf ("\ n"); continue;} Memset (Dp,0,sizeof (DP)); memset (vis,-1,sizeof (VIS)); for (i=0;i<len;i++) dp[i][i]=1;for (i=len-1;i>=0;i--) for (j=i+1;j<len;j++)         {dp[i][j]=dp[i+1][j]+1;            At first it was self-matching, that is, adding a vis[i][j]=i; if (judge (I,J))//If the end-to-end match, this special processing {if (Dp[i][j]>dp[i+1][j-1]) {dp[i][j]=dp[i+1][j-1];vis[i][j]=-1;}} for (int k=i+1;k<j;k++)//i~j, there is a match betweenif (judge (I,k)) {if (Dp[i][j]>dp[i][k]+dp[k+1][j]) {dp[i][j]=dp[i][k]+dp[k+1][j];vis[i][j]=k;}}} Print (0,len-1);p rintf ("\ n");} return 0;}

POJ 1141 Brackets Sequence (interval dp)