题意:给出一个模式串和一个主串,找出模式串在主串中第一次匹配的位置,不存在匹配时输出-1
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
思路:KMP,当有匹配时输出起点位置即可
注意点:无
以下为AC代码:
Run ID | Submit Time | Judge Status | Pro.ID | Exe.Time | Exe.Memory | Code Len. | Language | Author |
13036442 | 2015-03-05 19:40:56 | Accepted | 1711 | 1248MS | 12912K | 2616 B | G++ | luminous11 |
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <algorithm>
//#include <unordered_map>
//#include <unordered_set>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )
#define RDS(s) scanf ( "%s", s )
#define PIL(a) printf ( "%d\n", a )
#define PIIL(a,b) printf ( "%d %d\n", a, b )
#define PIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )
#define PSL(s) printf ( "%s\n", s )
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
const double eps = 1e-10;
const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
struct node{
int x, y, cnt;
node(){}
node( int _x, int _y ) : x(_x), y(_y) {}
node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
};
int pre[1000010];
int x[1000010];
int y[1000010];
void preKMP ( int x[], int m )
{
int i, j;
j = pre[0] = -1;
i = 0;
while ( i < m ){
while ( j != -1 && x[i] != x[j] ) j = pre[j];
pre[++i] = ++j;
//if ( x[++i] == x[++j] ) pre[i] = pre[j];
//else pre[i] = j;
}
}
int kmp ( int x[], int m, int y[], int n )
{
int i, j;
int ans = 0;
preKMP ( x, m );
i = j = 0;
while ( i < n ){
while ( j != -1 && x[j] != y[i] ) j = pre[j];
//cout << x[j] << ' ' << y[i] << ' ' << i << ' ' << j << endl;
i ++;
j ++;
if ( j >= m ){
return i - j + 1;
}
}
return -1;
}
int main()
{
int T;
RDI ( T );
while ( T -- ){
int m, n;
clr ( pre, 0 );
clr ( x, 0 );
clr ( y, 0 );
RDII ( n, m );
REP ( i, 0, n - 1 ) RDI ( x[i] );
REP ( i, 0, m - 1 ) RDI ( y[i] );
PIL ( kmp ( y, m, x, n ) );
}
return 0;
}