HDOJ 1711 Number Sequence

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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 IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor130364422015-03-05 19:40:56Accepted17111248MS12912K2616 BG++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;}

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