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.

**题意:**问每个0的四个方位里有几个是有1的,求出数量的和。
**思路:**题目要求宽松,基本想怎么写就怎么写...

/** @Date    : 2016-11-20-17.59

* @Author : Lweleth (SoungEarlf@gmail.com)

* @Link : https://github.com/

* @Version :

*/

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

//#include<bits/stdc++.h>

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+2000;



vector<int>row[1010],col[1010];

int mp[1010][1010];



int main()

{

int n, m;

cin >> n >> m;

for(int i = 1; i <= n; i++)

{

for(int j = 1; j <= m; j++)

{

scanf("%d", &mp[i][j]);

if(mp[i][j] == 1)

{

row[i].push_back(j);

col[j].push_back(i);

}

}

}

LL ans = 0;

for(int i = 1; i <= n; i++)

{

for(int j = 1; j <= m; j++)

{

if(mp[i][j])

continue;

int l = 0, r = 0;

for(int k = 0; k < row[i].size(); k++)

{

if(row[i][k] > j && r == 0)

r = 1;

if(row[i][k] < j && l == 0)

l = 1;

if((l == r && l == 1 ) || (r == 1 && row[i][k] > i))

break;

}

ans += l + r;

l = r = 0;

for(int k = 0; k < col[j].size(); k++)

{

if(col[j][k] > i && r == 0)

r = 1;

if(col[j][k] < i && l == 0)

l = 1;

if((l == r && l == 1 )|| (r == 1 && col[j][k] > j))

break;

}

ans += l + r;

}

}

printf("%lld\n", ans);





return 0;

}


Codeforces Round #380 (Div. 2)/729B Spotlights 水题的更多相关文章

  1. Codeforces Round #185 (Div. 2) B. Archer 水题

    B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B D ...

  2. Codeforces Round #360 (Div. 2) A. Opponents 水题

    A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents ...

  3. Codeforces Round #190 (Div. 2) 水果俩水题

    后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟 ...

  4. Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)解题报告

    对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespac ...

  5. Codeforces Round #340 (Div. 2) B. Chocolate 水题

    B. Chocolate 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co Bob loves everyt ...

  6. Codeforces Round #340 (Div. 2) A. Elephant 水题

    A. Elephant 题目连接: http://www.codeforces.com/contest/617/problem/A Descriptionww.co An elephant decid ...

  7. Codeforces Round #340 (Div. 2) D. Polyline 水题

    D. Polyline 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co There are three p ...

  8. Codeforces Round #338 (Div. 2) A. Bulbs 水题

    A. Bulbs 题目连接: http://www.codeforces.com/contest/615/problem/A Description Vasya wants to turn on Ch ...

  9. Codeforces Round #282 (Div. 1) A. Treasure 水题

    A. Treasure Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/494/problem/A ...

随机推荐

  1. HDU 1403 Longest Common Substring(后缀自动机——附讲解 or 后缀数组)

    Description Given two strings, you have to tell the length of the Longest Common Substring of them. ...

  2. mysql 启动报错

    之前用我这个机器做mysql的测试来,今天启动准备搭建一套线上的主从,结果起不来了... 错误日志: ;InnoDB: End of page dump 170807 11:37:02 InnoDB: ...

  3. Ubuntu16.04修改IP

    首先用root用户登陆,然后输入你root的密码.如下图:   然后编辑interfaces文件,该文件位于/etc/network/下,执行如下命令: vim /etc/network/interf ...

  4. 【Linux】CentOS安装redis

    CENTOS7下安装REDIS 安装完成之后使用:redis-cli命令连接,如图: 提示:/var/run/redis_6379.pid exists, process is already run ...

  5. 【转】how can i build fast

    http://blog.csdn.net/pcliuguangtao/article/details/5830860

  6. 获得system32等系统文件权限

    SYSTEM是至高无上的超级管理员帐户.默认情况下,我们无法直接在登录对话框上以SYSTEM帐户的身份登录到Windows桌面环境.实际上SYSTEM帐户早就已经“盘踞”在系统中了.根据http:// ...

  7. Java模块化开发

    包配置, 静态资源, 视图解析器, 数据库,

  8. 初入py

    1.下载工具sublime 我的网盘下载地址:https://pan.baidu.com/s/18-U1ZSg_zHoSAqUuvXj_PQ 直接解压即可 2.配置py27 在新建的文件里面编辑并保存 ...

  9. [剑指Offer] 54.字符流中的第一个不重复的字符

    题目描述 请实现一个函数用来找出字符流中第一个只出现一次的字符.例如,当从字符流中只读出前两个字符"go"时,第一个只出现一次的字符是"g".当从该字符流中读出 ...

  10. Foundation框架—时间输出格式NSDateFormatter

    在开发iOS程序时,有时候需要将时间格式调整成自己希望的格式,这个时候我们可以用NSDateFormatter类来处理.下面来介绍NSDateFormatter的常用属性和API:  1.常用属性 @ ...