CodeForces 699B——One Bomb（预处理，暴力枚举）

最新推荐文章于 2019-04-17 15:40:27 发布

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本篇介绍了一道算法题目，目标是在给定的地图上通过放置并引爆一颗炸弹来清除所有墙壁。详细解析了如何通过行和列的预先计算来判断是否能够实现目标，以及如何找到放置炸弹的最佳位置。

C - One Bomb

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

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

题意：有一些墙要炸。炸弹可以炸自己这行和这列，问一个炸弹是不是能炸光。

思路：预处理一下每行和每列有多少炸弹，并且记录一下炸弹总数。

枚举每个位置。如果这个位置不是炸弹，那么就要满足该行和该列炸弹总数等于总炸弹数。如果是炸弹，那就加起来减一等于总炸弹数。

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <set>


using namespace std;

#define MAXN 1010
#define LEN 1000000
#define INF 1e9+7
#define MODE 1000000
typedef long long ll;


char a[MAXN][MAXN];
int dp1[MAXN];
int dp2[MAXN];
set <int> res;
int main()
{
    int n,m;
    int sum=0;
    scanf("%d%d",&n,&m);
    getchar();
    for(int i=1;i<=n;i++)
    {
        scanf("%s",a[i]+1);
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
    {
        if(a[i][j]=='*')
        {
            dp1[i]++;
            dp2[j]++;
            sum++;
        }
    }
    int ok=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]=='*')
            {
                if(dp1[i]+dp2[j]-1==sum)
                {
                    printf("YES\n%d %d\n",i,j);
                    return 0;
                }
            }
            else
            {
                if(dp1[i]+dp2[j]==sum){
                    printf("YES\n%d %d\n",i,j);
                    return 0;
                }
            }
        }
    }
    if(!ok)
        printf("NO\n");
}