poj 1159 Palindrome 动态规划的三种解法

最长回文子串→最长公共子序列

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 5000    ;

string a,b;
int n;
int dp[2][maxn+5];
int main()
{
while(~scanf("%d",&n))
{
cin>>a;
b=a;
reverse(b.begin(),b.end());
a=" "+a;
b=" "+b;
int now=0;
int pre=1;
memset(dp[now],0,sizeof dp[now]);

for(int i=1;i<=n;i++)
{
now^=1;
pre^=1;
dp[now][0]=0;
for(int j=1;j<=n;j++)
{
if(a[i]==b[j]) dp[now][j]=dp[pre][j-1]+1;

else
{
dp[now][j]=max(dp[now][j-1],dp[pre][j]);
}
}
for(int j=0;j<=n;j++)
{
dp[pre][j]=dp[now][j];
}
}
printf("%d\n",n-dp[now]
);

}

return 0;
}

最长回文子串->dp[le][ri]:区间[le,ri]的最长回文子串长度

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 5000    ;

string a;
int n;
int dp[3][maxn+5];
int main()
{
while(~scanf("%d",&n))
{
cin>>a;
a=" "+a;
int now=2,pre=1,prepre=0;
for(int le=1;le<=n;le++)
{
dp[2][le]=1;
}
memset(dp[1],0,sizeof dp[1]);
for(int add=1;add<n;add++)
{
now=(now+1)%3;
pre=(pre+1)%3;
prepre=(prepre+1)%3;

for(int le=1;le+add<=n;le++)
{
int ri=le+add;
dp[now][le]=0;
if(a[le]==a[ri])
{
dp[now][le]=dp[prepre][le+1]+2;
}
else
{
dp[now][le]=max(dp[now][le],dp[pre][le]);
dp[now][le]=max(dp[now][le],dp[pre][le+1]);
}
}
}
printf("%d\n",n-dp[now][1]);

}

return 0;
}

dp[le][ri]:区间[le,ri]最少需要添加多少字符

1.未用滚动数组，会超内存

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 5000   ;
int n;
char s[maxn+10];
int dp[maxn+5][maxn+5];
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",s+1);
for(int i=1;i<=n;i++)
{
dp[i][i]=0;
dp[i][i-1]=0;
}
dp[n+1]
=0;

for(int add=1;add<n;add++)
{
for(int le=1;le+add<=n;le++)
{
int ri=le+add;
dp[le][ri]=add+1;
if(s[le]==s[ri])  dp[le][ri]=min(dp[le][ri],dp[le+1][ri-1]);
dp[le][ri]=min(dp[le][ri],dp[le+1][ri]+1  );
dp[le][ri]=min(dp[le][ri],dp[le][ri-1]+1  );

}
}
printf("%d\n",dp[1]
);

}

return 0;
}

2.滚动数组

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int maxn= 5000   ;
int n;
char s[maxn+10];
int dp[3][maxn+5];
int main()
{
while(~scanf("%d",&n))
{
scanf("%s",s+1);

memset(dp,0,sizeof dp);

int now=2,pre=1,prepre=0;

for(int add=1;add<n;add++)
{
now= (now+1)%3;
pre= (pre+1)%3;
prepre= (prepre+1)%3;

for(int le=1;le+add<=n;le++)
{
int ri=le+add;
dp[now][le]=add+1;
if(s[le]==s[ri])  dp[now][le]=min(dp[now   ][le],dp[ prepre  ][le+1]);
dp[now][le]=min(dp[now][le],dp[pre][le+1]+1  );
dp[now][le]=min(dp[now][le],dp[pre][le]+1  );

}
}
printf("%d\n",dp[now][1]);

}

return 0;
}