2016 北邮暑期训练3-C题（CodeForces 699B One Bomb）白痴题

最新推荐文章于 2018-11-25 14:56:00 发布

原创最新推荐文章于 2018-11-25 14:56:00 发布 · 733 阅读

0 ·

CC 4.0 BY-SA版权

another oj 专栏收录该内容

62 篇文章

订阅专栏

本文介绍了一个编程问题，即在一个由墙壁(*)和空地(.)组成的矩形区域中，通过放置一枚炸弹来炸毁所有墙壁的方法。炸弹引爆时可以清除其所在行和列的所有墙壁。文章提供了一种解决方案，并附带了实现该逻辑的C++代码。

Description
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.

Sample Input
Input
3 4
.*..
….
.*..
Output
YES
1 2
Input
3 3
..*
.*.
*..
Output
NO
Input
6 5
..*..
..*..

..*..
..*..
..*..
Output
YES
3 3

题目大意：

输入一个矩阵，其中*表示墙，.表示路，有一个炸弹，引爆时能炸掉所在行和列的墙，问能不能放置一个炸弹炸毁所有的墙？如果能，输出坐标。

题目分析：

用两个数组分别记录每一行和每一列的墙数，然后暴力枚举每一个点就行了。在LC里这也就是easy难度。

ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
  int m,n;
  char cell[1005][1005];
  int row[1005],col[1005];
  int wall=0;
  scanf("%d %d",&n,&m);
  for(int i=0;i<n;i++) {
    scanf("%s",cell[i]);
    row[i]=0;
    for(int j=0;j<m;j++)
      if(cell[i][j]=='*') {
        row[i]++;
        wall++;
      }
  }
  for(int j=0;j<m;j++) {
    col[j]=0;
    for(int i=0;i<n;i++) {
      if(cell[i][j]=='*')
        col[j]++;
    }
  }
  for(int i=0;i<n;i++) {
    for(int j=0;j<m;j++) {
      int bombs=row[i]+col[j];
      if(cell[i][j]=='*')
        bombs--;
      if(bombs==wall) {
        printf("YES\n%d %d\n",i+1,j+1);
        return 0;
      }
    }
  }
  printf("NO\n");
}