KMP算法题集

模板

caioj 1177 KMP模板

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e6 + 10;
const int MAXM = 1e3 + 10;
char a[MAXN], b[MAXM]; 
int next[MAXM], lena, lenb; 

void get_next()
{
	next[1] = 0; int j = 0;
	_for(i, 2, lenb)
	{
		while(j > 0 && b[j + 1] != b[i]) j = next[j];
		if(b[j + 1] == b[i]) j++;
		next[i] = j;	
	} 
}

void kmp()
{
	int j = 0;
	_for(i, 1, lena)
	{
		while(j > 0 && (j == lenb || b[j + 1] != a[i]))  j = next[j]; 
		if(b[j + 1] == a[i]) j++;
		if(j == lenb) { printf("%d %d
", i - lenb + 1, i); return; }	
	}
	puts("NO");
}

int main()
{
	scanf("%s%s", a + 1, b + 1);
	lena = strlen(a + 1); lenb = strlen(b + 1);
	get_next();
	kmp();
	return 0;
}

caioj 1460: 【KMP】字符串匹配

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e6 + 10;
char a[MAXN], b[MAXN]; 
int next[MAXN], lena, lenb;

void get_next()
{
	next[1] = 0; int j = 0;
	_for(i, 2, lena)
	{
		while(j > 0 && a[j + 1] != a[i]) j = next[j];
		if(a[j + 1] == a[i]) j++;
		next[i] = j;
	}
}

int kmp()
{
	int res = 0, j = 0;
	_for(i, 1, lenb)
	{
		while(j > 0 && a[j + 1] != b[i]) j = next[j];
		if(a[j + 1] == b[i]) j++;
		if(j == lena) { res++; j = next[j]; }	
	}	
	return res;
} 

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%s%s", a + 1, b + 1);
		lena = strlen(a + 1); lenb = strlen(b + 1); 
		get_next();
		printf("%d
", kmp());
	}
	return 0;
}

重复子串结论

有一个结论。

对于字符串S[1~i],如果i % (i - next[i]) == 0，那么这个字符串就由很多个重复的子串构成（形如abababab）

每个循环节等于S[1~i-next[i]]，循环节的个数为i / (i - next[i])

这个结论很好证明，用笔画一下就可以发现这个性质

caioj 1457: 【KMP】重复的子串

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e6 + 10;
char a[MAXN]; 
int next[MAXN], lena; 

void get_next()
{
	next[1] = 0; int j = 0;
	_for(i, 2, lena)
	{
		while(j > 0 && a[j + 1] != a[i]) j = next[j];
		if(a[j + 1] == a[i]) j++;
		next[i] = j;	
	} 
}

int main()
{
	while(scanf("%s", a + 1))
	{
		if(a[1] == '.') break;
		lena = strlen(a + 1);
		get_next();
		if(lena % (lena - next[lena]) == 0) 
			printf("%d
", lena / (lena - next[lena]));
		else puts("1");
	}
	return 0;
}

caioj 1458: 【KMP】判断循环段位置

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e6 + 10;
char a[MAXN]; 
int next[MAXN], lena; 

void get_next()
{
	next[1] = 0; int j = 0;
	_for(i, 2, lena)
	{
		while(j > 0 && a[j + 1] != a[i]) j = next[j];
		if(a[j + 1] == a[i]) j++;
		next[i] = j;	
	} 
}

int main()
{
	scanf("%s", a + 1);
	lena = strlen(a + 1);
	get_next();
	_for(i, 2, lena)
		if(i % (i - next[i]) == 0 && i / (i - next[i]) > 1)
			printf("%d %d
", i, i / (i - next[i]));
	return 0;
}

next数组应用

caioj 1459: 【KMP】所有前缀等于后缀的情况

#include<bits/stdc++.h>
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e6 + 10;
char a[MAXN]; 
int next[MAXN], lena; 

void get_next()
{
	next[1] = 0; int j = 0;
	_for(i, 2, lena)
	{
		while(j > 0 && a[j + 1] != a[i]) j = next[j];
		if(a[j + 1] == a[i]) j++;
		next[i] = j;	
	} 
}

int main()
{
	while(~scanf("%s", a + 1))
	{
		lena = strlen(a + 1);
		get_next();
		int j = lena;
		stack<int> s;
		while(j) s.push(j), j = next[j];
		while(!s.empty()) printf("%d ", s.top()), s.pop();
		puts("");
	}
	return 0;
}

综合题

poj 2185

这道题不知道为什么一直A不了。先放着

#include<cstdio>
#include<queue>
#include<cstring> 
#include<algorithm> 
#define REP(i, a, b) for(register int i = (a); i < (b); i++)
#define _for(i, a, b) for(register int i = (a); i <= (b); i++)
using namespace std;

const int MAXN = 1e4 + 10;
const int MAXM = 75 + 10;
char s[MAXN][MAXM], t[MAXN];
int next[MAXN], f[MAXN];
int n, m, r, c, j;

int get_r() //一行看作一个字符，很牛逼。 
{
	next[1] = 0;
	for(int i = 2, j = 0; i <= n; i++)
	{
		if(j > 0 && strcmp(s[j + 1], s[i])) j = next[j];
		if(!strcmp(s[j + 1], s[i])) j++;
		next[i] = j;
	}
	return n - next[n];
}

int get_c()
{
	memset(f, 0, sizeof(f));
	_for(i, 1, n)
		_for(len, 1, m)
			REP(j, 0, m)
			{
				if(s[i][j] != s[i][j % len]) break; //这个操作要学学 
				if(j == m - 1) f[len]++;
			}	
	_for(i, 1, m) //用桶这个思路很妙 
		if(f[i] == n)
			return i;
}

int main()
{
	scanf("%d%d", &n, &m); 
	_for(i, 1, n) scanf("%s", s[i]);
	printf("%d
", get_r() * get_c());
	return 0;
}