WustOJ 1575 Gingers and Mints(快速幂 + dfs )

最新推荐文章于 2022-01-29 15:54:08 发布

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最新推荐文章于 2022-01-29 15:54:08 发布

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原文链接：http://www.cnblogs.com/ZP-Better/p/5405839.html

本文详细解析了GingersandMints算法题目的背景、输入输出格式、样例及解答思路。通过DFS求出联通块个数，利用数学知识进行快速幂运算，最终输出答案模1000000007。

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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;
}