Codeforces Round #363 (Div. 2) B 暴力

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.

 

题意：

n*m地图上有不定数量的炸弹，问你能否选一个引爆点引爆所有的炸弹（引爆点可以引爆该行和该列的炸弹，即十字形）

 

思路：

一开始瞎JB模拟，弄了半天也没弄出来，后来听别人说才恍然大悟，只要找到位置x使得x.row_bomb+x.col_bomb-(x=='*')==tot_bomb，就是引爆点（x位置上可能有炸弹，也可能没有）。

 

下面是代码（copy from Rank 1……）

#include <string>
#include <cstdio>

const int maxn=1010;

int n,m;

char mp[maxn][maxn];

int row[maxn],col[maxn],tol=0;

int main(){
    scanf("%d %d",&n,&m);

    for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);

    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(mp[i][j]=='*'){
                row[i]++;
                col[j]++;
                tol++;
            }
        }
    }

    bool Find=false;
    int x,y;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(row[i]+col[j]-(mp[i][j]=='*')==tol){
                x=i;
                y=j;
                Find=true;
            }
        }
        if(Find) break;
    }

    if(Find) printf("YES\n%d %d",x,y);
    else printf("NO");

    return 0;
}