Title: It is legal to give a sequence of parentheses and brackets to satisfy the three conditions in a question. When not satisfied, it is illegal to add a minimum number of parentheses to an illegal sequence to make it legal. Outputs the shortest legal sequence.
Topic Analysis: This is an example in the "Getting Started classic". If only the shortest sequence is very simple, the definition DP (I,J) represents the minimum number of characters to be added as a valid interval i~j.
DP (I,J) =DP (i+1,j-1) (I and J can match), DP (I,J) =min (DP (I,K) +DP (k+1,j)).
However, the output time is relatively slow. From left to right, select the best scheme recursive output (see the book Code) by comparison.
The code is as follows:
# include<iostream># include<cstdio># include<cstring># include<algorithm>using namespace Std;char p[105];int dp[105][105];const int inf=100000;bool match (int x,int y) {if (p[x]== ' (' &&p[y]== ') ') return true; if (p[x]== ' [' &&p[y]== '] ') return true; return false;} void DP () {int len=strlen (p); for (int l=1;l<=len;++l) {for (int i=0;i+l-1<len;++i) {int r=i+l-1; if (l==1) {dp[i][r]=1; Continue } if (l==2) {if (Match (i,r)) dp[i][r]=0; else dp[i][r]=2; Continue } Dp[i][r]=inf; if (Match (I,R)) dp[i][r]=dp[i+1][r-1]; for (int k=i;k<r;++k) {dp[i][r]=min (dp[i][r],dp[i][k]+dp[k+1][r]); }}}}void print (int i,int j) {if (i>j) return; if (i==j) {if (p[i]==' (' | | p[i]== ') printf ("()"); else printf ("[]"); return; } int ans=dp[i][j]; if (Match (I,J) &&ans==dp[i+1][j-1]) {printf ("%c", P[i]); Print (i+1,j-1); printf ("%c", P[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 () {int T; scanf ("%d", &t); GetChar (); while (t--) {GetChar (); Gets (p); int Len=strlen (p); if (len>0) {DP (); Print (0,len-1); } printf ("\ n"); if (T) printf ("\ n"); } return 0;}
UVA-1626 Brackets sequence (simple interval DP)