poj 3167（KMP+树状数组）

之前自己在做题的时候在网上找别人的题解，虽然当时理解，但时间一久就忘了。所以开个这个东西来记录自己的学习进程，方便自己的回顾，以及给他人提供题解。

 

开始冲击明年高二的省选！不再颓废！

 

好了，下面进入正题

 

 

 

Cow Patterns

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3144		Accepted: 1164

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.

题目大意：

稍稍神奇的KMP。

模式串和匹配串匹配的条件是：匹配串各个位置数字的大小关系与模式串相同。或者说：把匹配串离散化后与模式串相同。

思路：

在普通的KMP中，当时s[i..j]和t[k..l]已经匹配时，s[i..j+1]和t[k..l+1]匹配的条件是s[j+1]=t[l+1]；而在本题中，要求匹配串各个位置数字的大小关系与模式串相同，

那么s[i..j+1]和t[k..l+1]匹配的条件便是s[i..j+1]中小于（等于）s[j+1]的数的个数与t[k..l]中小于（等于）t[l+1]的数的个数相同。用一个树状数组来记录s[i..j]/t[k..l]中小于（等于）

s[j+1]/t[l+1]的数的个数即可

其他与普通KMP基本相同。

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

int m1[100011],m2[100011],next[100011];
int a[100011],b[100011],ans[100011],f[101];
int n,m,t,i,j,xzq,x;

int lowbit(int x)
{
    return x&-x;
}

int get(int k)
{
    int l;
    l=0;
    while(k>0){
        l+=f[k];
        k-=lowbit(k);
    }
    return l;
}

void add(int k,int v)
{
    while(k<=t){
        f[k]+=v;
        k+=lowbit(k);
    }
}

int main()
{
    scanf("%d%d%d",&m,&n,&t);
    for(i=1;i<=m;i++)scanf("%d",&b[i]);
    for(i=1;i<=n;i++)scanf("%d",&a[i]);
    for(i=1;i<=n;i++){
        m1[i]=get(a[i]-1);
        m2[i]=get(a[i])-m1[i];
        add(a[i],1);
    }
    memset(f,0,sizeof(f));
    x=0;
    for(i=2;i<=n;i++){
        while(x!=0&&(get(a[i]-1)!=m1[x+1]||get(a[i])-get(a[i]-1)!=m2[x+1])){
            for(j=i-x;j<=i-1-next[x];j++)add(a[j],-1);
            x=next[x];
        }
        if(get(a[i]-1)==m1[x+1]&&get(a[i])-get(a[i]-1)==m2[x+1]){
            add(a[i],1);
            x++;
        }
        next[i]=x;
    }
    x=0;
    memset(f,0,sizeof(f));
    for(i=1;i<=m;i++){
        while(x!=0&&(get(b[i]-1)!=m1[x+1]||get(b[i])-get(b[i]-1)!=m2[x+1])){
            for(j=i-x;j<=i-1-next[x];j++)add(b[j],-1);
            x=next[x];
        }
        if(get(b[i]-1)==m1[x+1]&&get(b[i])-get(b[i]-1)==m2[x+1]){
            add(b[i],1);
            x++;
        }
        if(x==n){
            ans[++xzq]=i-n+1;
            for(j=i-x+1;j<=i-next[x];j++)add(b[j],-1);
            x=next[x];
        }
    }
    printf("%d
",xzq);
    for(i=1;i<=xzq;i++)printf("%d
",ans[i]);
}