B. One Bomb
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

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

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples
input
3 4
.*..
....
.*..
output
YES
1 2
input
3 3
..*
.*.
*..
output
NO
input
6 5
..*..
..*..
*****
..*..
..*..
..*..
output
YES
3 3
#include<stdio.h>
#include<iostream>
#include<map>
using namespace std;
char ma[][];
int main(){
int n,m;
int x=-;
int y=-;
int hang[];
int lie[];
int cnt=;
scanf("%d%d",&m,&n);
for(int i=;i<m;i++) hang[i]=;
for(int j=;j<n;j++) lie[j]=;
for(int i=;i<m;i++){
scanf("%s",ma[i]);
}
for(int i=;i<m;i++){
for(int j=;j<n;j++){
if(ma[i][j]=='*'){
hang[i]++;
lie[j]++;
cnt++;
}
} }
for(int i=;i<m;i++){
for(int j=;j<n;j++){
int tmp=hang[i]+lie[j];
if(ma[i][j]=='*') tmp--;
if(tmp==cnt){
printf("YES\n%d %d",i+,j+);
return ;
}
}
}
printf("NO\n");
return ;
}

分别统计行和列中到墙的个数,如果某个 行和列中包含全部到墙,则为答案。

cf #363 b的更多相关文章

  1. cf #363 d

      time limit per test 2 seconds memory limit per test 256 megabytes input standard input output stan ...

  2. cf #363 c

    C. Vacations time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  3. cf #363 a

    A. Launch of C time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...

  4. ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'

    凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...

  5. cf之路,1,Codeforces Round #345 (Div. 2)

     cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....   ...

  6. cf Round 613

    A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...

  7. ARC下OC对象和CF对象之间的桥接(bridge)

    在开发iOS应用程序时我们有时会用到Core Foundation对象简称CF,例如Core Graphics.Core Text,并且我们可能需要将CF对象和OC对象进行互相转化,我们知道,ARC环 ...

  8. [Recommendation System] 推荐系统之协同过滤(CF)算法详解和实现

    1 集体智慧和协同过滤 1.1 什么是集体智慧(社会计算)? 集体智慧 (Collective Intelligence) 并不是 Web2.0 时代特有的,只是在 Web2.0 时代,大家在 Web ...

  9. CF memsql Start[c]UP 2.0 A

    CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...

随机推荐

  1. Spark IDEA 调试(反编译)

    1)以WordCount为例,具体代码如下: import org.apache.spark.SparkConf import org.apache.spark.SparkContext; impor ...

  2. dubbo安装(转载)

    1.   概述 ZooKeeper是Hadoop的正式子项目,它是一个针对大型分布式系统的可靠协调系统,提供的功能包括:配置维护.名字服务.分布式同步.组服务等.ZooKeeper的目标就是封装好复杂 ...

  3. How to determine what causes a particular wait type

      By: Paul Randal Posted on: March 18, 2014 6:55 pm   [Edit 2016: Check out my new resource – a comp ...

  4. Swift,结构体与类

    1.结构体(小的类就是用struct来写) struct arrow{ var x=0,y=0 } 2.类(常用) class a{ var a=10 var b=20 } var b=a() //实 ...

  5. Oracle导入本属于sys用户的表

    FlashBack Database后,将删除的数据导出时使用了system用户 exp system/oracle file=/home/oracle/test.dmp tables=sys.tes ...

  6. fedora25 采用二进制包安装mysql5.5.49

    #添加用户和组 groupadd mysql useradd -s /sbin/nologin -g mysql -M mysql /etc/passwd id mysql #安装依赖包 [root@ ...

  7. vue-resource的使用中需要注意的坑

    先看一段代码: export default { name: 'app', data() { return { articles: [] } }, created: function() { this ...

  8. HAWQ技术解析(十八) —— 问题排查

    (原文地址:http://hawq.incubator.apache.org/docs/userguide/2.1.0.0-incubating/troubleshooting/Troubleshoo ...

  9. 贯通Spark Streaming流计算框架的运行源码

    本章节内容: 一.在线动态计算分类最热门商品案例回顾 二.基于案例贯通Spark Streaming的运行源码 先看代码(源码场景:用户.用户的商品.商品的点击量排名,按商品.其点击量排名前三): p ...

  10. Kubernetes用户指南(一)--快速开始、使用k8s配置文件

    一.快速开始 1.启动一个简单的容器. 一旦在container中打包好应用并将其commit为image之后,你就可以将其部署在k8s集群上. 一个简单的nginx服务器例子: 先决条件:你需要拥有 ...