Description

A particular subgroup of K (1 <= K <= 25,000) of Farmer John’s cows likes to make trouble. When placed in a line, these troublemakers stand together in a particular order. In order to locate these troublemakers, FJ has lined up his N (1 <= N <= 100,000) cows. The cows will file past FJ into the barn, staying in order. FJ needs your help to locate suspicious blocks of K cows within this line that might potentially be the troublemaking cows. 

FJ distinguishes his cows by the number of spots 1..S on each cow’s coat (1 <= S <= 25). While not a perfect method, it serves his purposes. FJ does not remember the exact number of spots on each cow in the subgroup of troublemakers. He can, however, remember which cows in the group have the same number of spots, and which of any pair of cows has more spots (if the spot counts differ). He describes such a pattern with a sequence of K ranks in the range 1..S. For example, consider this sequence: 

      1 4 4 3 2 1

In this example, FJ is seeking a consecutive sequence of 6 cows from among his N cows in a line. Cows #1 and #6 in this sequence have the same number of spots (although this number is not necessarily 1) and they have the smallest number of spots of cows #1..#6 (since they are labeled as ‘1’). Cow #5 has the second-smallest number of spots, different from all the other cows #1..#6. Cows #2 and #3 have the same number of spots, and this number is the largest of all cows #1..#6. 

If the true count of spots for some sequence of cows is: 

 5 6 2 10 10 7 3 2 9

then only the subsequence 2 10 10 7 3 2 matches FJ’s pattern above. 

Please help FJ locate all the length-K subsequences in his line of cows that match his specified pattern.

Input

Line 1: Three space-separated integers: N, K, and S 

Lines 2..N+1: Line i+1 describes the number of spots on cow i. 

Lines N+2..N+K+1: Line i+N+1 describes pattern-rank slot i.

Output

Line 1: The number of indices, B, at which the pattern matches 

Lines 2..B+1: An index (in the range 1..N) of the starting location where the pattern matches.

Sample Input

9 6 10
5
6
2
10
10
7
3
2
9
1
4
4
3
2
1

Sample Output

1
3

Hint

Explanation of the sample: 

The sample input corresponds to the example given in the problem statement. 

There is only one match, at position 3 within FJ’s sequence of N cows.

Source

USACO 2005 December Gold

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 100010;
const int MAXM = 25010;
int a[MAXN];
int b[MAXN];
int n, m, s;
int as[MAXN][30];
int bs[MAXM][30];

void init() { //预处理，计算当前数包含多少相同值的数 （s<=25，所以这里直接开了数组计算）
    for(int i = 0; i < n; ++i) {
        if(i ==0) {
            for(int j = 1; j <= 25; ++j) as[i][j] = 0;
        } else {
            for(int j = 1; j <= 25; ++j) as[i][j] = as[i-1][j];
        }
        as[i][a[i]]++;
    }
    for(int i = 0; i < m; ++i) {
        if(i ==0) {
            for(int j = 1; j <= 25; ++j) bs[i][j] = 0;
        } else {
            for(int j = 1; j <= 25; ++j) bs[i][j] = bs[i-1][j];
        }
        bs[i][b[i]]++;
    }
}

int next[MAXM];
void kmp_pre() {
    int i, j;
    j = next[0] = -1;
    i = 0;
    while(i < m) {
        int t11 = 0, t12 = 0, t21 = 0, t22 = 0;
        for(int k = 1; k < b[i]; k++) { //比b[i]小的数有多少
            if(i-j > 0) t11 += bs[i][k] - bs[i-j-1][k];
            else t11 += bs[i][k];
        }
        if(i-j > 0) t12 = bs[i][b[i]] - bs[i-j-1][b[i]];  //等于b[i]的数有多少
        else t12 = bs[i][b[i]];

        for(int k = 1; k < b[j]; k++) {  //比b[j]小的数有多少
            t21 += bs[j][k];
        }
        t22 = bs[j][b[j]];  //等于b[j]的数有多少
        if(j == -1 || (t11 == t21 && t12 == t22)) {
            next[++i] = ++j;
        } else {
            j = next[j];
        }
    }
}

vector<int>ans;
void kmp() {
    ans.clear();
    int i, j;
    kmp_pre();
    i = j = 0;
    while(i < n) {
        int t11 = 0, t12 = 0, t21 = 0, t22 = 0;
        for(int k = 1; k < a[i]; k++) {
            if(i-j > 0) t11 += as[i][k] - as[i-j-1][k];
            else t11 += as[i][k];
        }
        if(i-j > 0) t12 = as[i][a[i]] - as[i-j-1][a[i]];
        else t12 = as[i][a[i]];

        for(int k = 1; k < b[j]; k++) {
            t21 += bs[j][k];
        }
        t22 = bs[j][b[j]];
        if(j == -1 || (t11 == t21 && t12 == t22)) { //两个位置的数具有同等偏向关系取决于两个数的 比自身值小 和 等于自身值 这两个量相等
            i++; j++;
            if(j >= m) {
                ans.push_back(i-m+1);
                j = next[j];
            }
        } else {
            j = next[j];
        }
    }
}

int main()
{
    //freopen("out.txt", "w", stdout);
    //freopen("in.txt", "r", stdin);
    while(scanf("%d%d%d", &n, &m, &s) == 3) {
        for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
        for(int i = 0; i < m; ++i) scanf("%d", &b[i]);
        init();
        kmp();
        printf("%d\n", ans.size());
        for(int i = 0; i < ans.size(); ++i) {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}