uva 10739

dp 只有三个操作  当str[i] != str[j] 时 dp(i, j) = min(dp(i+1, j), dp(i+1, j-1), dp(i, j-1))

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstring>
#include <algorithm>
using namespace std;

char str[1010];
int f[1005][1005];
int n;
int dp(int a, int b)
{
    if(f[a][b] != -1)
        return f[a][b];
    if(a >= b)
        return f[a][b] = 0;
    if(str[a] == str[b])
        f[a][b] = dp(a+1, b-1);
    else
        f[a][b] = min(dp(a+1, b-1), min(dp(a+1, b), dp(a, b-1)))+1;
    return f[a][b];
}
int main()
{
    int t, ca = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str);
        int len = strlen(str);
        memset(f, -1, sizeof(f));
        printf("Case %d: %d\n", ca++, dp(0, len-1));
    }
    return 0;
}