CF330 A. Cakeminator 大水题

A. Cakeminator

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rectangular cake, represented as an r × c grid. Each cell either has an evil strawberry, or is empty. For example, a 3 × 4 cake may look as follows:

The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.

Please output the maximum number of cake cells that the cakeminator can eat.

Input

The first line contains two integers r and c (2 ≤ r, c ≤ 10), denoting the number of rows and the number of columns of the cake. The next r lines each contains c characters — the j-th character of the i-th line denotes the content of the cell at row i and column j, and is either one of these:

• '.' character denotes a cake cell with no evil strawberry;
• 'S' character denotes a cake cell with an evil strawberry.

Output

Output the maximum number of cake cells that the cakeminator can eat.

Sample test(s)

input

3 4S.........S.

output

8

Note

For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).

题目解决：

如果一个点有草莓的话，就会限制掉那对应的一行与对应的列，当一个点所处的坐标（x,y）行x,列y都被限制时。就是此点不能去除，不计算该点。所以解法出来了。

代码：

/* * @author ipqhjjybj * @date  20130720 * */#include <cstdio>#include <cstdlib>#include <algorithm>#include <iostream>#include <cstring>using namespace std;#define ll long long#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define clr(x,k) memset(x,k,sizeof(x))char s[100][100];int heng[100][100];int shu[100][100];int main(){    //freopen("330A.in","r",stdin);    int n,m;    clr(heng,0);    clr(shu,0);    scanf("%d %d",&n,&m);    getchar();    for(int i = 0;i < n;i++)        gets(s[i]);    for(int i = 0;i < n;i++){        for(int j = 0;j < m;j++){            if(s[i][j]=='S'){                for(int ii = 0;ii<m;ii++){                    heng[i][ii]=1;                }                for(int jj = 0;jj<n;jj++){                    shu[jj][j]=1;                }            }        }    }    int ans = 0;    for(int i = 0;i < n;i++)        for(int j = 0;j < m;j++){            if(!heng[i][j]||!shu[i][j]){                ans++;            }        }    printf("%d\n",ans);    return 0;}