KMP算法题集
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\n", 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;
}#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\n", 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
#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\n", lena / (lena - next[lena]));
else puts("1");
}
return 0;
}#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\n", i, i / (i - next[i]));
return 0;
}next数组应用
#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
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\n", get_r() * get_c());
return 0;
}
