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.
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
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
使用深度优先遍历,4个方向。如果遍历到则置为'0',只遍历一次。
#include <iostream>
#include <stdio.h>
using namespace std;
int M, N;
char map[60][60];
int dir2[4][2] = { {-1,0},{0,-1},{1,0},{0,1} };
int dir[11][4] = {
{1,1,0,0},{0,1,1,0},{1,0,0,1},{0,0,1,1},{0,1,0,1},{1,0,1,0},{1,1,1,0},{1,1,0,1},{1,0,1,1},{0,1,1,1},{1,1,1,1}
};
void dfs(int m,int n){
int index = map[m][n] - 'A';
map[m][n] = '0';
//cout << m << "<>" << n << endl;
for(int i=0; i<4; i++){
if(dir[index][i]){
int tempx = n + dir2[i][0];
int tempy = m + dir2[i][1];
if( tempx >=0 && tempx < N && tempy >=0
&& tempy < M && map[tempy][tempx] != '0'){
int tempIndex = map[tempy][tempx] - 'A';
int k = i+2;
if(k > 3)
k -= 4;
if(dir[tempIndex][k])
dfs(tempy, tempx);
}
}
}
}
int main() {
// cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
while(cin >>M >> N){
if(M <= 0 || N <= 0)
break;
for(int i=0; i<M; i++)
scanf("%s", map[i]);
int cnt = 0;
for(int i=0; i<M; i++){
for(int j=0; j<N; j++){
if(map[i][j] != '0'){
dfs(i,j); //i是行,j是列
cnt ++;
}
}
}
cout << cnt << endl;
}
return 0;
}
分享到:
相关推荐
ZOJ解题报告ZOJ解题报告ZOJ解题报告ZOJ解题报告
zoj题目简单归类zoj题目简单归类zoj题目简单归类
acm中zoj1002的可运行C++程序
包含了zoj700多道题目的源代码,在做题时可以参考
Problem Arrangement zoj 3777
ZOJ题目答案源码
学习ACM程序设计的朋友一定要看,这是训练必备的POJ ZOJ题目分类及解题思路
一个非常非常非常非常实用的zoj结题代码
zoj 1003 c语言的,要写这么多描述吗。。
浙大ZOJ题目分类,可以让你更方便快速锁定那你想要联系的题目,是自己快速提高·
本代码是zoj上AC的1951的代码,把双重循环简化为O(n),不过素数判断的改进还不够
ZOJ1805代码
zoj1027解题指南和代码,还不错,是学校培训给的。
ZOJ题解集合-截至2835。共1244个文件,C/C++,有重复
zoj 题库 详细解答 解题代码 acm
zoj4041正确题解源代码,以及运行程序
zoj吐血制作,希望大家喜欢
大学ACM竞赛,ZOJ 1733 运用递归(优化)的方法。ac的代码。
能AC 通过的c++代码,包括zoj1002,1091,1789
zoj3464 Rugby Football测试数据