UVA 11404 plalidromic subsquence

The longest palindrome sequence can be obtained by solving the original string s and the LCS of the reverse serial RV, because the palindrome string is required to be dp[i][j, and the length is preserved.

Requires a minimum dictionary order, Dp[i][j] should represent the end of a palindrome string, so the boundary is a single character, that is, i+j=len+1.

The most troublesome part of the problem lies in the dictionary order, I wrote a comparison function, a bit violent (constant large).

It is also possible to reverse the definition, when the node is going to hold the string of that state (this defines a small constant when comparing the dictionary order)

#include <bits/stdc++.h>using namespacestd;#defineMP Make_pair#defineFi first#defineSe SecondConst intLEN = 1e3+5;CharS[len],rv[len];intDp[len][len];p air<int,int>Pre[len][len];CharVal[len][len];inlinevoidUpdata (intIintJintVCharCConstpair<int,int> &PRV) {Dp[i][j]=v; PRE[I][J]=PRV; VAL[I][J]=C;}ConstAuto Nil = MP (0,0);#defineDim (x) [x.fi][x.se]BOOLCmplex (pair<int,int> a,pair<int,int>b) {    //     while(A! = Nil && val[a.fi][a.se] = =val[b.fi][b.se]) {a= Pre Dim (a); b =Pre Dim (b); }    returnVal Dim (A) <Val Dim (b);}//#define LOCALintMain () {#ifdef LOCAL freopen ("In.txt","R", stdin);#endif     while(gets (s)) {intLen =strlen (s);  for(inti =0; i < Len; i++) {Rv[len-1-I.] =S[i]; }        intHD1,HD2,VL =0;  for(inti =1; I <= Len; i++){            intj = len+1-i; DP[I][J]=1; VAL[I][J] = s[i-1]; PRE[I][J] =Nil; if(Dp[i][j] > VL | | (Dp[i][j] = = VL && Cmplex (MP (I,J), MP (HD1,HD2)))) {//VL =Dp[i][j]; HD1= i; HD2 =J; }             for(intK =1; K < J; k++) Dp[i][k] =0;  forJ <= Len; j + +){                if(s[i-1] = = rv[j-1]) {updata (i,j,dp[i-1][j-1]+2, s[i-1],make_pair (I-1, J-1)); if(Dp[i][j] > VL | | (Dp[i][j] = = VL && Cmplex (MP (I,J), MP (HD1,HD2)))) {//VL =Dp[i][j]; HD1= i; HD2 =J; }                }Else {                    if(dp[i-1][J] > dp[i][j-1] || (dp[i-1][J] = = dp[i][j-1] && Cmplex (MP (i-1, j), MP (i,j-1)) ) ){//Updata (i,j,dp[i-1][j],val[i-1][j],pre[i-1][j]); }Else{updata (i,j,dp[i][j-1],val[i][j-1],pre[i][j-1]); }                }            }        }        intPV = (vl+1) >>1, LN =VL; Auto U=MP (HD1,HD2);  for(inti =0; I < PV; i++) {S[i]=val[u.fi][u.se]; U=pre[u.fi][u.se]; } S[ln]=' /';  for(inti = PV; i < ln; i++) {S[i]= s[ln-1-i];    } puts (s); }    return 0;}

UVA 11404 plalidromic subsquence