UVA 10739 String to Palindrome

UVA_10739

    我们用f[i][j]表示字符串从第i位到第j位构成回文串最少所需的操作数，如果b[i]==b[j]，f[i][j]=f[i+1][j-1]，否则f[i][j]=min{f[i][j-1],f[i+1][j],f[i+1][j-1]}+1。

#include<stdio.h>
#include<string.h>
#define MAXD 1010
#define INF 1000000000
char b[MAXD];
int f[MAXD][MAXD];
int dp(int x, int y)
{
    int t, min = INF;
    if(f[x][y] != -1)
        return f[x][y];
    if(x > y)
        return 0;
    if(b[x] == b[y])
    {
        t = dp(x + 1, y - 1);
        if(t < min)
            min = t;
    }
    else
    {
        t = dp(x + 1, y);
        if(t + 1 < min)
            min = t + 1;
        t = dp(x, y - 1);
        if(t + 1 < min)
            min = t + 1;
        t = dp(x + 1, y - 1);
        if(t + 1 < min)
            min = t + 1;
    }
    return f[x][y] = min;
}
void solve()
{
    int k, res;
    memset(f, -1, sizeof(f));
    scanf("%s", b);
    k = strlen(b) - 1;
    res = dp(0, k);
    printf("%d\n", res);
}
int main()
{
    int t, tt;
    scanf("%d", &t);
    for(tt = 0; tt < t; tt ++)
    {
        printf("Case %d: ", tt + 1);
        solve();
    }
    return 0;
}