【CodeForces】699A - One Bomb(思维)
2016-07-20 11:02
513 查看
点击打开题目
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
output
input
output
input
output
记录每一横行和每一竖行有多少墙,然后两个for嵌套扫描出每一点能排掉多少墙,全部排掉就得到了答案。
代码如下:
#include <cstdio>
int main()
{
int w,h;
int b;
char map[1000+11][1000+11];
while (~scanf ("%d %d",&h,&w))
{
getchar();
int ww[1011] = {0};
int hh[1011] = {0};
b = 0; //墙数
for (int i = 1 ; i <= h ; i++)
{
for (int j = 1 ; j <= w ; j++)
{
scanf ("%c",&map[i][j]);
if (map[i][j] == '*')
{
b++;
hh[i]++;
ww[j]++;
}
}
getchar();
}
int flag = false;
int ans_x,ans_y;
for (int i = 1 ; i <= h ; i++)
{
for (int j = 1 ; j <= w ; j++)
{
int ant = hh[i] + ww[j];
if (map[i][j] == '*')
ant--;
if (ant == b)
{
flag = true;
ans_x = i;
ans_y = j;
break;
}
}
if (flag)
break;
}
if (flag)
printf ("YES\n%d %d\n",ans_x,ans_y);
else
printf ("NO\n");
}
return 0;
}
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
记录每一横行和每一竖行有多少墙,然后两个for嵌套扫描出每一点能排掉多少墙,全部排掉就得到了答案。
代码如下:
#include <cstdio>
int main()
{
int w,h;
int b;
char map[1000+11][1000+11];
while (~scanf ("%d %d",&h,&w))
{
getchar();
int ww[1011] = {0};
int hh[1011] = {0};
b = 0; //墙数
for (int i = 1 ; i <= h ; i++)
{
for (int j = 1 ; j <= w ; j++)
{
scanf ("%c",&map[i][j]);
if (map[i][j] == '*')
{
b++;
hh[i]++;
ww[j]++;
}
}
getchar();
}
int flag = false;
int ans_x,ans_y;
for (int i = 1 ; i <= h ; i++)
{
for (int j = 1 ; j <= w ; j++)
{
int ant = hh[i] + ww[j];
if (map[i][j] == '*')
ant--;
if (ant == b)
{
flag = true;
ans_x = i;
ans_y = j;
break;
}
}
if (flag)
break;
}
if (flag)
printf ("YES\n%d %d\n",ans_x,ans_y);
else
printf ("NO\n");
}
return 0;
}
相关文章推荐
- Android SQLite操作之“SQLiteOpenHelper”
- yarn资源调度设置
- Android SearchView 自定义SearchIcon和字体颜色大小
- Java List排序
- 支付宝服务窗API接口开发php版本
- 状态(State)模式
- PHP- 数字转汉字
- Vim快捷键整理
- NBU所用端口统计
- Hdfs将服务器的整个文件夹拷贝到Linux本地
- 关于ARM9协处理器CP15及MCR和MRC指令
- 利用POI抽取word中的图片并保存在文件中
- 神器推荐:应用之星,零代码就能制作炫酷的H5页面
- LeetCode Binary Tree Level Order Traversal I.II
- UVA 10250-The Other Two Trees
- uc/os-iii学习笔记-时间管理
- MFC修改对话框及控件背景颜色 - 很轻松!
- 设计模式(五) 原型模式
- 苹果企业号-通过网页下载应用,部署应用分发服务器
- Service Intent must be explicit错误