Codeforces Round #515 (Div. 3) B. Heaters【贪心区间合并细节】

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本文详细解析了Codeforces B. Heaters题目，介绍了一种使用贪心算法解决加热器覆盖问题的方法，通过尽可能扩展右端点来找到最小数量的加热器，以覆盖所有房间。

任意门：http://codeforces.com/contest/1066/problem/B

Vova's house is an array consisting of $n$ 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 $i$ -th element of the array is $1$ if there is a heater in the position $i$ , otherwise the $i$ -th element of the array is $0$ .

Each heater has a value $r$ ( $r$ is the same for all heaters). This value means that the heater at the position $p o s$ can warm up all the elements in range $[p o s - r + 1; p o s + r - 1]$ .

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 $n = 6$ , $r = 2$ and heaters are at positions $2$ and $5$ , then Vova can warm up the whole house if he switches all the heaters in the house on (then the first $3$ elements will be warmed up by the first heater and the last $3$ 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 $n$ and $r$ ( $1 \leq n, r \leq 1000$ ) — the number of elements in the array and the value of heaters.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ( $0 \leq a_{i} \leq 1$ ) — 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 $2$ warms up elements $[1; 3]$ , the heater at the position $3$ warms up elements $[2, 4]$ and the heater at the position $6$ warms up elements $[5; 6]$ so the answer is $3$ .

In the second example the heater at the position $1$ warms up elements $[1; 3]$ and the heater at the position $5$ warms up elements $[3; 5]$ so the answer is $2$ .

In the third example there are no heaters so the answer is -1.

In the fourth example the heater at the position $3$ warms up elements $[1; 5]$ , the heater at the position $6$ warms up elements $[4; 8]$ and the heater at the position $10$ warms up elements $[8; 10]$ so the answer is $3$ .

题意概括：

N 个房间，标记为 1 的房间有加热器，加热范围【 i-r+1， i+r-1】；求使用最少的加热器加热所有房间。

解题思路:

这个贪心并不难想，顺序遍历，尽可能的扩展右端点。

但wa了好多发，主要是区间连接的细节问题。

ans_ed 为已覆盖集合的最右端的下一个。

看代码吧，心好凉。

AC code：

 1 #include <bits/stdc++.h>
 2 #define LL long long
 3 #define INF 0x3f3f3f3f
 4 using namespace std;
 5 const int MAXN = 1e3+1;
 6 
 7 int a[MAXN], l[MAXN], r[MAXN], cnt;
 8 int ans_ed, now_st, now_ed; //当前已选集合能覆盖的最右端的下一个！！！下标！！！，当前结点能覆盖的左端，当前结点能覆盖的右端
 9 int pre_index, now_index;   //前一个可加的下标，当前可加的下标
10 int N, R;
11 int ans;
12 bool chch;
13 
14 int main()
15 {
16     scanf("%d%d", &N, &R);
17     bool flag = true;
18     cnt = 0;
19     for(int i = 1; i <= N; i++){
20         scanf("%d", &a[i]);
21         if(a[i] == 1){
22             flag = false;
23             //nxt[++cnt] = i;
24             cnt++;
25             l[cnt] = i-R+1; //当前可加热的左端
26             r[cnt] = i+R-1; //当前可加热的右端
27         }
28     }
29     //see see
30     //for(int i = 1; i <= cnt; i++) printf("%d ", nxt[i]);
31     //puts("");
32     if(flag) printf("-1\n");    //没有加热器
33     else{
34         chch = false;
35         ans = 0;
36         ans_ed = 1;             //答案最右端初始化为1
37         int k = 1;              //从头开始顺序遍历加热器
38         bool book = false;
39         do{
40             if(l[k] < ans_ed)   //当前加热器k可连接前一个加热区，说明该点可用
41             {
42                 book = true;    //标记有保留考虑项
43                 pre_index = k;  //记录当前下标
44                 k++;            //讨论下一个，如果下一个可加，则当前这个不加
45                 if(k == cnt+1){   //如果当前这个为最后一个
46                     if(ans_ed < N+1){ //当前未全覆盖
47                         ans++;
48                         ans_ed = r[k-1]+1;      //注意加1
49                     }
50                 }
51                 continue;
52             }
53 
54             if(book){           //说明前一个可选，但是未选
55                 if(l[k] > ans_ed){         //当前这个不可以加，说明上一个要加
56                     now_index = pre_index;
57                 }
58                 else now_index = k;          //当前这个可加
59                 ans++;
60                 ans_ed = r[now_index]+1;
61                 if(now_index == pre_index) --k;
62                 book = false;
63             }
64             else if(l[k] > ans_ed) //当前加热器不能连接前一个加热区，说明出现断层，但是是顺序遍历下来的，说明无解
65             {
66                 chch = true;break;
67             }
68             else{                      //无缝连接，选！
69                 ans++;                //答案加一
70                 ans_ed = r[k]+1;       //更新集合最右端
71             }
72             k++;                   //讨论下一个；
73         }while(k <= cnt);
74         if(ans_ed <= N || ans == 0) chch = true;
75         if(chch) printf("-1\n");
76         else printf("%d\n", ans);
77     }
78     return 0;
79 }