2015 ACM Amman Collegiate Programming Contest L .Alternating Strings II
This problem is the same as problem D but with different constraints!
Bahosain has a strange habit. He writes his daily notes in binary! His little brother also has a strange habit. He
hates seeing alternating patterns of 0s and 1s or strings longer than
K
. Therefore, whenever Bahosain writes a
binary note on a paper, his little brother cuts it at night with scissors in order to make each part of it not
longer than
K
and not alternating.
After several years of suffering from cutting binary notes with scissors, Bahosain decided to cut his notes in
such a way that his brother won’t touch them.
Given a binary string of length
N
, find the minimum number of cuts Bahosain has to make such that each of
the resulting strings are NOT alternating and contains no more than
K
bits.
A binary string is considered alternating only if each bit after the first one is different from the one before it.
For example: strings 110, 0110 and 010100 are not alternating, while 101 and 01 are alternating strings.
A string with one character is not an alternating string.
Input
The first line of input contains
T (1 ≤ T ≤ 128)
that represents the number of test cases.
The first line of each test case contains two integers:
N
and
K (1 ≤ K ≤ N ≤ 100,000)
, where
N
is the length
of the string and
K
is the maximum length of a resulting string.
The next line contains a string of
N
bits (0 or 1).
Output
For each test case, print a single line with the minimum number of cuts.
Sample Input
4
6 3
111000
5 2
11010
3 3
110
3 3
101
Sample Output
3
0
2
解法：（find）我们记录以i结尾的且是交替串的最长长度为pre，可以看到在计算f[i]时，有以下几种情况：
1.s[i] == s[i - 1] 这时候没有以i结尾交替出现的串，重置pre=1，f[i] = min{f[i - k], ..., f[i - 1]} + 1 }
2.s[i] != s[i - 1] 这时候pre++,由于最后一个与前一个形成交替串，所以需要比较pre与k的关系:
a. 若pre >= k 或 pre == i，着说明连续出现的交替串长度>k，或者前i个全是01串，那么此时f[i]只能由f[i - 1]转移（相当于在i-1与i之间切一刀）
b.否则的话pre + 1这个串（[i - pre - 1, i])一定不是交替串,那么此时[i - k, i - pre - 1]这个区间的所有点都可以转移到i（最后pre+1个不是交替串，那么最后pre+1+x个一定也不是），所以有f[i] = min{f[i - k], ..., f[i - pre - 1]} + 1，所以可以使用线段树查询区间最小值，复杂度NlogN
总结：对于10^5一般对应于NlogN的算法。对于区间查询类的dp，我们可以考虑用线段树来优化。线段树的id必须从1开始，区间可以从0-n。
AC：
#include <stdio.h>
#include <algorithm>
#include<iostream>
using namespace std;
struct s
{
int left;
int right;
int mi;
}tree[800008];
char s[800008];
void build(int id,int l,int r)
{
tree[id].left=l;
tree[id].right=r;
if(l==r)
{
tree[id].mi=0;
}
else{
int mid=(l+r)/2;
build(2*id,l,mid);
build(2*id+1,mid+1,r);
tree[id].mi = min(tree[id*2].mi,tree[2*id+1].mi);
}
}
void update(int id,int x,int y)
{
if(tree[id].left ==tree[id].right )
{
tree[id].mi =y;
}
else
{
int mid=(tree[id].left +tree[id].right )/2;
if(x>mid) update(2*id+1,x ,y);
else update(2*id,x,y);
tree[id].mi =min(tree[2*id].mi,tree[2*id+1].mi);
}
}
int query(int id,int l,int r)
{
if(tree[id].left >=l&&tree[id].right <=r)
return tree[id].mi;
else
{
int mid=(tree[id].left +tree[id].right )/2;
if(l>mid) return query(id*2+1,l,r);
else if(r<=mid) return query(id*2,l,r);
else return min(query(id*2+1,l,r),query(id*2,l,r));
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int pre,val,n,k;
pre=1;
scanf("%d%d",&n,&k);
scanf("%s",s+1);
build(1,0,n);
update(1,1,1);
for(int i=2;i<=n;i++)
{
if(s[i]==s[i-1])
{
pre=1;
val=query(1,max(i-k,0),i-1);
}
else {
pre++;
if(pre>=k||pre==i)
{
val=query(1,i-1,i-1);
}
else val=query(1,max(i-k,0),i-1-pre);
}
update(1,i,val+1);
}
printf("%d\n",query(1,n,n)-1);
}
}