HDU 2253 Longest Common Subsequence Again

AC代码是网上找来的，我没读懂。留待以后来看

#include <cstdio>
#include <cstring>

#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;
	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, 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();
}