B. One Bomb (#363 Div.2)

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

題意：你有一個炸彈，可以放在任意位置并炸掉該位置所在的行和列上的墻“*”，求一顆炸彈是否能炸掉所有的墻。

我們可以將每行和每列的墻數分別存入數組x[]和y[]。每當a[i][j]的位置是"*"時，x[i]++,y[j]++;最后再將x[i]+y[j]與總墻數sum比較。

注意：當炸彈所在點為“*”時，需x[i]+y[j]-1;

附AC代碼：

 1 #include<iostream>
 2 #include<cstring>
 3 
 4 using namespace std;
 5 
 6 int N,M,ans;
 7 int a[1002][1002],x[1002],y[1002];
 8 char s[200010];
 9 
10 void process(){
11     scanf("%d %d",&N,&M);
12     int cnt = 0;
13     for(int i=1; i<=N; i++){
14         scanf("%s",s);
15         for(int j=1; j<=M; j++){
16             if(s[j-1] == '*'){
17                 x[i]++;
18                 y[j]++;
19                 cnt++;
20                 a[i][j] = 1;
21             }
22         }
23     }
24     for(int i=1; i<=N; i++){
25         for(int j=1; j<=M; j++){
26             int t;
27             t = x[i]+y[j];
28             if(a[i][j] == 1) t--;//重復一個 
29             if(t == cnt){
30                 printf("YES\n%d %d\n",i,j);
31                 return;
32             }
33         }
34     }
35     printf("NO\n");
36 }
37 
38 int main(){
39     process(); 
40     return 0;
41 }

轉載于:https://www.cnblogs.com/Kiven5197/p/5687391.html

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