ZOJ 2412 Farm Irrigation(DFS 条件通讯块)

ZOJ 2412 Farm Irrigation(DFS 条件通讯块)

大家好,又见面了,我是全栈君,今天给大家准备了Idea注册码。

意甲冠军  两个农田管内可直接连接到壳体  他们将能够共享一个水源   有11种农田  管道的位置高于一定  一个农田矩阵  问至少须要多少水源

DFS的连通块问题  两个相邻农田的管道能够直接连接的话他们就属于一个连通块  题目就是求连通块个数 

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 55;
char mat[N][N];
int type[11][4] = {   //相应11种水管类型 按顺时针方向有管道的为1否则为0
    {1, 0, 0, 1}, {1, 1, 0, 0}, {0, 0, 1, 1}, {0, 1, 1, 0},
    {1, 0, 1, 0}, {0, 1, 0, 1}, {1, 1, 0, 1}, {1, 0, 1, 1},
    {0, 1, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}
};

int dfs(int r, int c)
{
    int cur = mat[r][c] - 'A';
    if(cur < 0 || cur > 10) return 0;
    mat[r][c] = 0;           //标记已经訪问 0-'A'是小于0的
    int up = mat[r - 1][c] - 'A', dw = mat[r + 1][c] - 'A',
        le = mat[r][c - 1] - 'A', ri = mat[r][c + 1] - 'A';  //4个相邻块的管道类型
    if(up > -1 && up < 11 && type[up][2] && type[cur][0])  dfs(r - 1, c);
    if(dw > -1 && dw < 11 && type[dw][0] && type[cur][2])  dfs(r + 1, c);
    if(le > -1 && le < 11 && type[le][1] && type[cur][3])  dfs(r, c - 1);
    if(ri > -1 && ri < 11 && type[ri][3] && type[cur][1])  dfs(r, c + 1);
    return 1;     //一个连通块中仅仅有一个返回1
}

int main()
{
    int n, m, ans;
    while (scanf("%d%d", &n, &m), n >= 0)
    {
        ans = 0;
        memset(mat, 0, sizeof(mat));
        for(int i = 1; i <= n; ++i)
            scanf("%s", mat[i] + 1);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                ans += dfs(i, j);
        printf("%d\n", ans);
    }
    return 0;
}


Farm Irrigation



Time Limit: 2 Seconds      Memory Limit: 65536 KB


Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.


ZOJ 2412 Farm Irrigation(DFS 条件通讯块)
Figure 1

Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map

ADC
FJK
IHE

then the water pipes are distributed like


ZOJ 2412 Farm Irrigation(DFS 条件通讯块)
Figure 2

Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.

Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?

Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.

Input

There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of ‘A’ to ‘K’, denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.

Output

For each test case, output in one line the least number of wellsprings needed.

Sample Input

2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

Sample Output

2
3



版权声明:本文博客原创文章,博客,未经同意,不得转载。

版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 举报,一经查实,本站将立刻删除。

发布者:全栈程序员-用户IM,转载请注明出处:https://javaforall.cn/117484.html原文链接:https://javaforall.cn

【正版授权,激活自己账号】: Jetbrains全家桶Ide使用,1年售后保障,每天仅需1毛

【官方授权 正版激活】: 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...

(0)
blank

相关推荐

发表回复

您的电子邮箱地址不会被公开。

关注全栈程序员社区公众号