[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。

View Code

#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;

}