> #include <iostream>
> using namespace std;
> #include <string.h>
> #define N 5001
> int n;
> char s[N];
> short f[N][N];
> int solv(char s[N],int i,int j)
> {
> 	if(f[i][j])
> 		return f[i][j];
> 	if(i >= j)
> 		return f[i][j] = 0;
> 	if(s[i] == s[j])
> 		return f[i][j] = solv(s,i+1,j-1);
> 	if(solv(s,i,j-1) < solv(s,i+1,j))
> 		return f[i][j] = solv(s,i,j-1) + 1;
> 	return f[i][j] = solv(s,i+1,j) + 1;
> }
> int main(void)
> {
> 	scanf("%d",&n);
> 	scanf("%s",s);
> 	memset(f,0,sizeof(f));
> 	printf("%d\n",solv(s,0,strlen(s)-1));
> 	return 0;
> }

• Re:动态规划代码如下，注意用short，否则超内存 (12) Heng_Xing 2011-08-11 18:13:40
  • Re:感觉用欧拉路做，可是每个节点只能经过一次？这怎么搞 (34) JustName 2011-08-11 18:17:52