poj 3167 Cow Patterns（kmp）

Cow Patterns

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3127 Accepted: 1153

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 1056210107329144321

Sample Output

13

题意：告诉你n个数组成的串，和一个等级序列，问你符合这个等级序列的子串个数，和这些子串的起始位置。

思路：

如果不是符合等级，而是完全符合的话，直接用kmp就可以搞定。但是，这题的话，next数组怎么求呢？

我们知道，在求next数组的时候是用递推的：

if(p[i] == p[next[i]]) next[i+1] = next[i]+1;

本题在什么情况下p[i] == p[next[i]]？ 当且仅当p[i]在p[k](i-next[i]<=k<=i-1)间的等级 == 
p[next[i]]在p[k](0<=k<=next[i]-1)间的等级时才相等。
即如图：
开二维数组sum[i][25]来预处理记录从第0~i位，25内个数字出现的次数，就可以推出某个区间比p[k]小的数有多少个了！（当然，和p[k]相等的个数也可以求出来，就是比p[k]+1小的减去比p[k]小的）！
#include <iostream>#include <cstdio>#include <map>#include <vector>using namespace std;const int maxn1 = 100100;const int maxn2 = 25010;const int maxn3 = 27;int Rank[maxn2] , spot[maxn1];map<int , int> mp;int N , K , S , Next[maxn2];int sumS[maxn1][maxn3] , sumR[maxn2][maxn3] , cnt;vector<int> ans;void initial(){    for(int i = 0; i < maxn1; i++) spot[i] = 0;    for(int i = 0; i < maxn2; i++) Rank[i] = 0 , Next[i] = 0;    //for(int i = 0; i < maxn3; i++){        //for(int j = 0; j < maxn1; j++) sumS[j][i] = 0;        //for(int j = 0; j < maxn2; j++) sumR[j][i] = 0;   // }//初始化这里用G++交超时，C++1907MS。去掉G++907MS。- -！...    mp.clear();    ans.clear();    cnt = 1;}void readcase(){    for(int i = 0; i < N; i++){        scanf("%d" , &spot[i]);    }    for(int i =0; i < K; i++){        scanf("%d" , &Rank[i]);        mp[Rank[i]] = 0;    }    for(map<int,int>::iterator it = mp.begin(); it != mp.end(); it++) it->second = cnt++;    for(int i = 0; i < K; i++){        Rank[i] = mp[Rank[i]];    }}void get_sum(){    sumS[1][spot[0]] = 1;    sumR[1][Rank[0]] = 1;    for(int i = 1; i < N; i++){        for(int j = 0; j < maxn3; j++) sumS[i+1][j] = sumS[i][j];        sumS[i+1][spot[i]]++;    }    for(int i = 1; i < K; i++){        for(int j = 0; j < maxn3; j++) sumR[i+1][j] = sumR[i][j];        sumR[i+1][Rank[i]]++;    }}int getSpot_status(int l , int r , int num){    int sta = 0;    for(int i = 0; i < num; i++){        if(sumS[r][i]-sumS[l][i]>0) sta++;    }    return sta;}int getRank_status(int l , int r , int num){    int sta = 0;    for(int i = 0; i < num; i++){        if(sumR[r][i]-sumR[l][i] > 0) sta++;    }    return sta;}void getNext(){    Next[0] = -1;    for(int i = 0; i < K; i++){        int next = Next[i];        while(next >= 0 && !(getRank_status(0 , next , Rank[next]) == getRank_status(i-next , i , Rank[i]) && getRank_status(0 , next , Rank[next]+1) == getRank_status(i-next , i , Rank[i]+1))){            next = Next[next];        }        Next[i+1] = next+1;    }}void kmp(){    getNext();    /*for(int i = 0; i < K; i++){        cout << Next[i] << " ";    }    cout << endl;*/    int r = 0 , s = 0;    while(s < N){        if(getRank_status(0 , r , Rank[r]) == getSpot_status(s-r , s , spot[s])&&getRank_status(0 , r , Rank[r]+1) == getSpot_status(s-r , s , spot[s]+1)){            s++;            r++;        }else{            r = Next[r];            if(r < 0){                s++;                r = 0;            }        }        if(r >= K){            r = Next[r];            ans.push_back(s-K+1);        }    }}void computing(){    get_sum();    kmp();    printf("%d\n" , ans.size());    for(int i = 0; i < ans.size(); i++) printf("%d\n" , ans[i]);}int main(){    while(scanf("%d%d%d" , &N , &K , &S) != EOF){        initial();        readcase();        computing();    }    return 0;}