题目描述
Polycarp and his friends want to visit a new restaurant. The restaurant has n tables arranged along a straight line. People are already sitting at some tables. The tables are numbered from 1 to n in the order from left to right. The state of the restaurant is described by a string of length n which contains characters “1” (the table is occupied) and “0” (the table is empty).
Restaurant rules prohibit people to sit at a distance of k or less from each other. That is, if a person sits at the table number i, then all tables with numbers from i−k to i+k (except for the i-th) should be free. In other words, the absolute difference of the numbers of any two occupied tables must be strictly greater than k.
For example, if n=8 and k=2, then:
strings “10010001”, “10000010”, “00000000”, “00100000” satisfy the rules of the restaurant;
strings “10100100”, “10011001”, “11111111” do not satisfy to the rules of the restaurant, since each of them has a pair of “1” with a distance less than or equal to k=2.
In particular, if the state of the restaurant is described by a string without “1” or a string with one “1”, then the requirement of the restaurant is satisfied.
You are given a binary string s that describes the current state of the restaurant. It is guaranteed that the rules of the restaurant are satisfied for the string s.
Find the maximum number of free tables that you can occupy so as not to violate the rules of the restaurant. Formally, what is the maximum number of “0” that can be replaced by “1” such that the requirement will still be satisfied?
For example, if n=6, k=1, s= “100010”, then the answer to the problem will be 1, since only the table at position 3 can be occupied such that the rules are still satisfied.
Input
The first line contains a single integer t (1≤t≤104) — the number of test cases in the test. Then t test cases follow.
Each test case starts with a line containing two integers n and k (1≤k≤n≤2⋅105) — the number of tables in the restaurant and the minimum allowed distance between two people.
The second line of each test case contains a binary string s of length n consisting of “0” and “1” — a description of the free and occupied tables in the restaurant. The given string satisfy to the rules of the restaurant — the difference between indices of any two “1” is more than k.
The sum of n for all test cases in one test does not exceed 2⋅105.
Output
For each test case output one integer — the number of tables that you can occupy so as not to violate the rules of the restaurant. If additional tables cannot be taken, then, obviously, you need to output 0.
Example
input
6
6 1
100010
6 2
000000
5 1
10101
3 1
001
2 2
00
1 1
0
output
1
2
0
1
1
1
Note
The first test case is explained in the statement.
In the second test case, the answer is 2, since you can choose the first and the sixth table.
In the third test case, you cannot take any free table without violating the rules of the restaurant.
题目大意
给定01字串，若相邻的1之间0的个数大于等于k,是有效的社交距离。给定一个01字串，将其中的0变成1,还能保持有效的社交距离，求最多的变换次数。
题目分析
用一个pos[]数组表示座位的状态，pos[i]=true表示第i个位子可以坐人。
将字符串s正向遍历一遍。如果s[i]=1，那么让pos[i , i+k]=false。
将字符串s反向遍历一遍。如果s[i]=1，那么让pos[i-k , i]=false。
最后剩下的就是能坐人的位置了，不过注意在在最后统计空位的时候，还是要注意间隔。
代码如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=2e5+5;
bool pos[N];
int main()
{
int t;
cin>>t;
while(t--)
{
memset(pos,true,sizeof pos); //初始化
int n,k;
string s;
cin>>n>>k;
cin>>s;
for(int i=0;i<s.size();i++) //正向遍历
{
if(s[i]=='1') //s[i]==1,则将i———i+k的位置的pos赋成false
{
pos[i]=false;
for(int j=0;j<k&&i<s.size();j++)
{
i++;
pos[i]=false;
}
}
}
for(int i=s.size()-1;i>=0;i--)//反向遍历
{
if(s[i]=='1') //s[i]==1,则将i-k———i的位置的pos赋成false
{
pos[i]=false;
for(int j=0;j<k&&i>0;j++)
{
i--;
pos[i]=false;
}
}
}
int ans=0; //统计空位
for(int i=0;i<s.size();i++)
{
if(!pos[i]) continue;
ans++; //当找到某个位置可以坐人时，往后跳k位再继续统计
for(int j=0;j<k&&i<s.size();j++)
{
i++;
pos[i]=false;
}
}
cout<<ans<<endl;
}
return 0;
}