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 32char 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();}