1575: Gingers and Mints
Time Limit: 1 Sec Memory Limit: 128 MB 64bit IO Format: %lld
Submitted: 24 Accepted: 13
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Description
fcbruce
owns a farmland, the farmland has n * m grids. Some of the grids are
stones, rich soil is the rest. fcbruce wanna plant gingers and mints on
his farmland, and each plant could occupy area as large as possible. If
two grids share the same edge, they can be connected to the same area.
fcbruce is an odd boy, he wanna plant gingers, which odd numbers of
areas are gingers, and the rest area, mints. Now he want to know the
number of the ways he could plant.
Input
The first line of input is an integer T (T < 100), means there are T test cases.
For each test case, the first line has two integers n, m (0 < n, m < 100).
For next n lines, each line has m characters, ‘N‘ for stone, ‘Y‘ for rich soil that is excellent for planting.
Output
For each test case, print the answer mod 1000000007 in one line.
Sample Input 
2 3 3 YNY YNN NYY 3 3 YYY YYY YYY
Sample Output
4 1
HINT
For the
first test case, there are 3 areas for planting. We marked them as A, B
and C. fcbruce can plant gingers on A, B, C or ABC. So there are 4 ways
to plant gingers and mints.
Source
武汉科技大学第二届移动互联网应用设计大赛(A类)暨华中地区程序设计竞赛专业组网络赛
Author
fcbruce
思路：
1.先用DFS求出联通块个数。
2.再根据数学知识 C{n,1} + C{n,3} + C{n,5} + ... = 2^(n-1) 用快速幂求出。
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 200
char pic[maxn][maxn];
int vis[maxn][maxn];
int dx[4] = {-1,1,0,0};
int dy[4] = {0,0,-1,1};
int n,m;
void dfs(int x, int y)
{
for(int i = 0; i < 4; i++)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(!vis[nx][ny] && pic[nx][ny] == ‘Y‘ && nx >= 0 && nx < n && ny >= 0 && ny < m)
{
vis[nx][ny] = 1;
dfs(nx,ny);
}
}
}
int pow_mod(int a,int n,int m)
{
if(n==0)
return 1;
int x=pow_mod(a,n/2,m);
long long ans=(long long)x*x%m;
if(n%2==1)
ans=ans*a%m;
return (int )ans;
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
memset(vis, 0, sizeof vis);
for(int i = 0; i < n; i++)
scanf("%s", pic[i]);
int cnt = 0;
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
if(!vis[i][j] && pic[i][j] == ‘Y‘)
{
vis[i][j] = 1;
cnt++;
dfs(i,j);
}
}
printf("%d\n",pow_mod(2,cnt-1,1000000007));
}
return 0;
}