Longest Common Subsequence Again HDU 2253 LCS+位压缩

Everyone knows LCS problem: 

For example: 
str1 = "GCTAT" 
str2 = "CGATTA" 

the longest common subsequence is "GTT" of them.So you should print "3". 

I'm sure you can easily sovle it.So the challenge is coming again.(^_^) 

InputFor each case,there are two lines.one is string A,the other is string B. 
the string Process to end of file.Each string will have at most 30000 characters. 
All the characters are in upper-case. 
OutputFor each case, output the maximum-length common subsequence. 
Sample Input

GCTAT
CGATTA 

ABCFBC
ABFCAB

Sample Output

3
4
题目的意思就是LCS谁都会套模板了，现在要搞你，直接套TLE，所以加个位压缩大法盘它

#include <stdio.h>
#include <string.h>

#define M 30005
#define SIZE 128
#define WORDMAX 3200
#define BIT 32

char s1[M], s2[M];
int nword;
unsigned int str[SIZE][WORDMAX];
unsigned int tmp1[WORDMAX], tmp2[WORDMAX];

void pre(int len)
{
    int i, j;
    memset(str, 0, sizeof(str));
    for(i = 0; i < len; i ++)
        str[s1[i]][i / BIT] |= 1 << (i % BIT);
}

void cal(unsigned int *a, unsigned int *b, char ch)
{
    int i, bottom = 1, top;
    unsigned int x, y;
    for(i = 0; i < nword; i ++)
    {
        y = a[i];
        x = y | str[ch][i];
        top = (y >> (BIT - 1)) & 1;
        y = (y << 1) | bottom;
        if(x < y) top = 1;
        b[i] = x & ((x - y) ^ x);
        bottom = top;
    }
}

int bitcnt(unsigned int *a)
{
    int i, j, res = 0, t;
    unsigned int b[5] = {0x55555555, 0x33333333, 0x0f0f0f0f, 0x00ff00ff, 0x0000ffff}, x;
    for(i = 0; i < nword; i ++)
    {
        x = a[i];
        t = 1;
        for(j = 0; j < 5; j ++, t <<= 1)
            x = (x & b[j]) + ((x >> t) & b[j]);
        res += x;
    }
    return res;
}

void process()
{
    int i, j, len1, len2;
    unsigned int *a, *b, *t;
    len1 = strlen(s1);
    len2 = strlen(s2);
    nword = (len1 + BIT - 1) / BIT;
    pre(len1);
    memset(tmp1, 0, sizeof(tmp1));
    a = &tmp1[0];
    b = &tmp2[0];
    for(i = 0; i < len2; i ++)
    {
        cal(a, b, s2[i]);
        t = a; a = b; b = t;
    }
    printf("%d\n", bitcnt(a));
}

int main()
{
    while(scanf("%s%s", s1, s2) != EOF)
        process();
    return 0;
}