B. Heaters
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Vova's house is an array consisting of nn elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The ii-th element of the array is 11 if there is a heater in the position ii, otherwise the ii-th element of the array is 00.
Each heater has a value rr (rr is the same for all heaters). This value means that the heater at the position pospos can warm up all the elements in range [pos−r+1;pos+r−1][pos−r+1;pos+r−1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n=6n=6, r=2r=2 and heaters are at positions 22 and 55, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 33 elements will be warmed up by the first heater and the last 33 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers nn and rr (1≤n,r≤10001≤n,r≤1000) — the number of elements in the array and the value of heaters.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤10≤ai≤1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
input
Copy
6 2 0 1 1 0 0 1
output
Copy
3
input
Copy
5 3 1 0 0 0 1
output
Copy
2
input
Copy
5 10 0 0 0 0 0
output
Copy
-1
input
Copy
10 3 0 0 1 1 0 1 0 0 0 1
output
Copy
3
Note
In the first example the heater at the position 22 warms up elements [1;3][1;3], the heater at the position 33 warms up elements [2,4][2,4] and the heater at the position 66 warms up elements [5;6][5;6] so the answer is 33.
In the second example the heater at the position 11 warms up elements [1;3][1;3] and the heater at the position 55 warms up elements [3;5][3;5] so the answer is 22.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 33 warms up elements [1;5][1;5], the heater at the position 66 warms up elements [4;8][4;8] and the heater at the position 1010 warms up elements [8;10][8;10] so the answer is 33.
题目大意：有n个房子排成一排，部分房子有着火点，有着火点的房子用1表示，没有的用0表示。着火的房子的覆盖范围为[pos - r + 1) ,(pos + r - 1)].，最少点燃多少个房子可以把所有的房子点燃。
题目大致思路：我们每遇到一个着火的房子我们就把他所能覆盖的范围的房子的值更新一下，更新的值为将其点燃的房子的覆盖距离r，然后在处理每一个房子的值，看有没有为0的点。
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000010;
int a[MAXN];
int main(){
int n,r;
while(~scanf("%d %d",&n,&r)){
int k;
memset(a,0,sizeof(a));
for(int i = 1; i <= n; i++){
scanf("%d",&k);
if(k == 1){
for(int j = max(1,i - r + 1); j <= min(n,i + r - 1); j++){
a[j] = max(a[j],i);
}
}
}
int i = 1,ans = 0;
bool flag = false;
while(i <= n){
if(a[i] == 0){
printf("-1\n");
flag = true;
break;
}
ans++;
i = a[i] + r;
}
if(!flag) printf("%d\n",ans);
}
return 0;
}