KMP

参考：  http://www.cnblogs.com/BeyondAnyTime/archive/2012/07/09/2583133.html

        http://blog.csdn.net/v_july_v/article/details/7041827#t3    这个较好

以hdu 1711 为模板

方法一

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 1e6+9;const int maxn1 = 1e4+9;int a[maxn], b[maxn1];int n, m;int next[maxn1];void Next(){    next[0] = -1;    for(int j = 1; j < m; j++)    {        int i = next[j-1];        while(b[j] != b[i+1] && i >= 0) i = next[i];//如果不相等，就如他们匹配不成功，如 1 1 2 1 1 4, 4和2不等        if(b[j] == b[i+1]) next[j] = i+1;//如同 1 1 2与1 1 4匹配，所以 要 i = next[i];        else next[j] = -1;    }}int kmp(){    Next();    int p = 0, s = 0;    while(p < m && s < n)    {        if(b[p] == a[s]) p++, s++;        else        {            if(p == 0) s++;            else p = next[p-1] + 1;        }    }    if(p < m) return -1;    return s-m;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        memset(next, 0, sizeof(next));        scanf("%d%d", &n, &m);        for(int i = 0; i < n; i++) scanf("%d", a+i);        for(int i = 0; i < m; i++) scanf("%d", b+i);        int w = kmp();        printf("%d\n", w==-1?-1:w+1);    }    return 0;}

方法二

#include <iostream>#include <cstdio>using namespace std;const int maxn = 1e6+9;const int maxn1 = 1e4+9;int a[maxn], b[maxn1], n, m;int nextval[maxn1];void get_nextval(){    int i = 0, j = -1;    nextval[0] = -1;    while(i < m)    {        if(j == -1 || b[i] == b[j])        {             if(b[++i] == b[++j]) nextval[i] = nextval[j];             else nextval[i] = j;        }        else j = nextval[j];    }}void kmp(){    get_nextval();    int i = 0, j = 0;    while(i < n && j < m)    {        if(j == -1 || a[i] == b[j]) ++i, ++j;        else j = nextval[j];    }    if(j == m) printf("%d\n", i-j+1);    else printf("-1\n");}int main(){    int t;    cin >> t;    while(t--)    {        scanf("%d%d", &n, &m);        for(int i = 0; i < n; i++) scanf("%d", &a[i]);        for(int i = 0; i < m; i++) scanf("%d", &b[i]);        kmp();    }    return 0;}