HDU 4749 Parade Show 相对匹配的KMP

题目来源：HDU 4749 Parade Show

题意：从长度为n的序列最多能找到多少个不重叠的的连续的子串 和长度位m的序列的相对大小相同 

思路：转化成KMP 把原来的字母相同换成一个检测是否相对大小相同的函数 具体参考了http://blog.csdn.net/no__stop/article/details/11949743

#include 
#include 
const int maxn = 100010;
int a[maxn], b[maxn], f[maxn];
int sum1[30][maxn], sum2[30][maxn];
int n, m, l;
bool ok1(int x, int y)
{
	int s1 = 0, s2 = 0;
	if(sum2[b[x]][x-1] != sum2[b[y]][y-1] - sum2[b[y]][y-x])
		return false;
	for(int i = 0; i < b[x]; i++)
		s1 += sum2[i][x-1];
	for(int i = 0; i < b[y]; i++)
	{
		s2 += sum2[i][y-1];
		s2 -= sum2[i][y-x];
	}
	if(s1 != s2)
		return false;
	return true;
}
bool ok2(int x, int y)
{
	int s1 = 0, s2 = 0;
	if(sum2[b[x]][x-1] != sum1[a[y]][y-1] - sum1[a[y]][y-x])
		return false;
	for(int i = 0; i < b[x]; i++)
		s1 += sum2[i][x-1];
	for(int i = 0; i < a[y]; i++)
	{
		s2 += sum1[i][y-1];
		s2 -= sum1[i][y-x];
	}
	if(s1 != s2)
		return false;
	return true;
}
void getFail()
{
	f[0] = f[1] = 0;
	for(int i = 2; i <= m; i++)
	{
		int j = f[i-1];
		while(j && !ok1(j+1, i))
			j = f[j];
		f[i] = j+1;
	}
}
void find()
{
	int cnt = 0;
	int j = 0;
	for(int i = 1; i <= n; i++)
	{
		while(j && !ok2(j+1, i))
			j = f[j];
		j++;
		if(j == m)
		{
			cnt++;
			j = 0;
		} 
	}
	printf("%d\n", cnt);
}

int main()
{
	while(scanf("%d %d %d", &n, &m, &l) != EOF)
	{
		memset(sum1, 0, sizeof(sum1));
		memset(sum2, 0, sizeof(sum2));
		for(int i = 1; i <= n; i++)
		{
			int x;
			scanf("%d", &x);
			a[i] = x;
			for(int j = 1; j <= l; j++)
				sum1[j][i] = sum1[j][i-1];
			sum1[x][i]++;
		}
		for(int i = 1; i <= m; i++)
		{
			int x;
			scanf("%d", &x);
			b[i] = x;
		
			for(int j = 1; j <= l; j++)
				sum2[j][i] = sum2[j][i-1];
			sum2[x][i]++;
		}
		getFail();
		find();
	}
	return 0;
}