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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 =1005;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=0; i<maxn-1; ++i)     dp[i][0].len=dp[0][i].len=0,dp[i][0].s=dp[0][i].s="";   while(scanf("%s",s1+1)==1){       int n= strlen(s1+1);       for(int i=1;i<=n; ++i)        dp[i][i].len=1,dp[i][i].s=s1[i];       for(int i=1; i<=n; ++i){             for(int j=i-1; j>=1; --j){                    dp[j][i].len=0;                    dp[j][i].s="";                  if(s1[i]==s1[j]){                      dp[j][i].len=dp[j+1][i-1].len+2;                      dp[j][i].s+=s1[j];                      dp[j][i].s+=dp[j+1][i-1].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+1][i]<dp[j][i-1])                           dp[j][i]=dp[j][i-1];                      else dp[j][i]=dp[j+1][i];                  }             }       }       cout<<dp[1][n].s<<endl;   }    return 0;}
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uva11404