[poj][3167][Cow Patterns]

题目：http://poj.org/problem?id=3167

题意：给一个模式串，按照模式串的大小对应关系，找出匹配串有相同大小对应关系的子串。利用指针，下标可以直接一一对应。

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

const int N = 100000+10;
const int M = 25000+10;
int n, k, s, at;
int a[N], b[M];
int d[30], low[M], high[M], path[M], ans[N];

bool cmp(int *aa, int *bb, int k)
{
    if (aa[d[bb[k]]] != aa[k]) return false;
    if (low[k]>=0&&aa[low[k]]>=aa[k]) return false;
    if (high[k]>=0&&aa[high[k]]<=aa[k]) return false;
    return true;
}

int main()
{
    //freopen("D:/a.txt", "r", stdin);
    scanf("%d%d%d", &n, &k, &s);
    for (int i=1; i<=s; i++)d[i]=-1;
    d[0] = d[s+1] = -2, at=0;
    for (int i=0; i<n; i++)
        scanf("%d", &a[i]);
    for (int i=0, j; i<k; i++)
    {
        scanf("%d", &b[i]);
        if (d[b[i]]==-1) d[b[i]]=i;
        for (j=b[i]-1; d[j]==-1; j--);
        low[i] = d[j];
        for (j=b[i]+1; d[j]==-1; j++);
        high[i] = d[j];
    }
    path[0] = -1;
    for (int i=1,j=-1; i<k; i++)
    {
        while (j>=0&&!cmp(b+(i-j-1),b,j+1))j=path[j];
        if (cmp(b+(i-j-1),b,j+1))j++;
        path[i] = j;
    }
    for (int i=0,j=-1; i<n; i++)
    {
        while (j>=0&&!cmp(a+(i-j-1),b,j+1))j=path[j];
        if (cmp(a+(i-j-1),b,j+1))j++;
        if (j+1 == k)
        {
            ans[at++] = i-j+1;
            j = path[j];
        }
    }
    printf("%d\n", at);
    for (int i=0; i<at; i++)
        printf("%d\n", ans[i]);
    return 0;
}