uva11404

这题说的是给了一个长度为n的字符串（1000）求最长回文子序列，并输出当str[i]==ste[j]时dp[i][j]=dp[i+1][i-1]+2 否则 dp[i][j]=Max(dp[j+1][i],dp[j][i-1]) 要强调一下这uva真是强大 每个后面都加一个string都不爆内存太厉害了

#include <iostream>
#include <string.h>
#include <algorithm>
#include <string>
#include <cstdio>
using namespace std;
const int maxn =;
struct point{
    int len;
    string s;
    bool operator <(const point &A)const{
         return len<A.len||(len==A.len&&s>A.s);
    }
}dp[maxn][maxn];
char s1[maxn],s2[maxn];
int main()
{
   for(int i=; i<maxn-; ++i)
     dp[i][].len=dp[][i].len=,dp[i][].s=dp[][i].s="";
   while(scanf("%s",s1+)==){
       int n= strlen(s1+);
       for(int i=;i<=n; ++i)
        dp[i][i].len=,dp[i][i].s=s1[i];
       for(int i=; i<=n; ++i){
             for(int j=i-; j>=; --j){
                    dp[j][i].len=;
                    dp[j][i].s="";
                  if(s1[i]==s1[j]){
                      dp[j][i].len=dp[j+][i-].len+;
                      dp[j][i].s+=s1[j];
                      dp[j][i].s+=dp[j+][i-].s;
                      dp[j][i].s+=s1[i];
                     // if(dp[j][i]<dp[j+1][i]) dp[j][i]=dp[j+1][i];
                      //if(dp[j][i]<dp[j][i-1]) dp[j][i]=dp[j][i-1];
                  }else{
                      if(dp[j+][i]<dp[j][i-])
                           dp[j][i]=dp[j][i-];
                      else dp[j][i]=dp[j+][i];
                  }
             }
       }
       cout<<dp[][n].s<<endl;
   }

    return ;
}