Codeforce 729B. Spotlights

最新推荐文章于 2022-05-26 19:50:53 发布

原创最新推荐文章于 2022-05-26 19:50:53 发布 · 574 阅读

0 ·

CC 4.0 BY-SA版权

思维专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一个关于剧场舞台上灯光布局的算法问题。通过分析舞台计划图，确定放置聚光灯的有效位置，确保不会直接照射到演员且至少能照亮一名演员的方向。文章提供了具体的实现思路和示例，帮助理解如何计算有效位置的数量。

Theater stage is a rectangular field of size n × m. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.

You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.

A position is good if two conditions hold:

there is no actor in the cell the spotlight is placed to;
there is at least one actor in the direction the spotlight projects.

Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the plan.

The next n lines contain m integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.

Output

Print one integer — the number of good positions for placing the spotlight.

Examples

input
2 4
0 1 0 0
1 0 1 0

output
9

input
4 4
0 0 0 0
1 0 0 1
0 1 1 0
0 1 0 0

output
20

Note

In the first example the following positions are good:

the (1, 1) cell and right direction;
the (1, 1) cell and down direction;
the (1, 3) cell and left direction;
the (1, 3) cell and down direction;
the (1, 4) cell and left direction;
the (2, 2) cell and left direction;
the (2, 2) cell and up direction;
the (2, 2) and right direction;
the (2, 4) cell and left direction.

Therefore, there are 9 good positions in this example.

把图的外面一层都设置成1,那么对于所有0的贡献都是4，然后再减去多算的就行了，多算的就是贡献为为1的（按每行每列来看），所以从头或尾最长连续0的个数就是要减去的；还有一种方法设置cnt记录1之前的0的个数，flag标记前面是否有1，然后每行每列统计即可；

AC代码：

#include<cstdio>

const int MAXN=1e3+11;
int G[MAXN][MAXN];

int main()
{
	int N,M;
	while(~scanf("%d%d",&N,&M)) {
		int ans=0;
		for(int i=0;i<N;++i) {
			for(int j=0;j<M;++j) {
				scanf("%d",&G[i][j]); 
				ans+=(!G[i][j]);
			}
		}
		ans*=4;
		for(int i=0;i<N;++i) {
		    int cnt=0;
		    int j=0;
		    while(!G[i][j]&&j<M) cnt++,++j;
		    j=M-1;
		    while(!G[i][j]&&j>=0) cnt++,--j;
		    ans-=cnt;
		}
		for(int i=0;i<M;++i) {
			int cnt=0;
			int j=0;
			while(!G[j][i]&&j<N) cnt++,++j;
			j=N-1;
			while(!G[j][i]&&j>=0) cnt++,--j;
			ans-=cnt;
		}
		printf("%d\n",ans);
	}
	return 0;
 }