Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2520 Accepted Submission(s):
1134
| 15MS | 1452K |
#include<stdio.h>
#include<string.h>
#define min(a,b) (a>b?b:a)
int dp[110][110],ans[110];
char str1[110],str2[110];
int main()
{
while(scanf("%s%s",str1+1,str2+1)!=EOF)
{
int i,j,k;
int len=strlen(str1+1);
memset(dp,0,sizeof(dp));
memset(ans,0,sizeof(ans));
for(i=1;i<=len;i++)
dp[i][i]=1;
for(i=len-1;i>=1;i--)//a串和b串一个也不相等时
{
for(j=i+1;j<=len;j++)
{
dp[i][j]=dp[i+1][j]+1;
for(k=i+1;k<=j;k++)
{
if(str2[i]==str2[k])
{
dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);
}
}
}
}
for(i=1;i<=len;i++)
{
ans[i]=dp[1][i];
if(str1[i]==str2[i])
{
ans[i]=ans[i-1];
}
else
{
for(j=1;j<i;j++)
{
ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
}
}
}
printf("%d\n",ans[len]);
}
}
HDOJ 题目2474 String painter(区间DP)
原文:http://www.cnblogs.com/Who-you/p/4856010.html