AK is an ACM competition finished all of the problems. AC This problem is one of the steps,so fight to AK。There are a number of graphics"_" and "#"，In the graph,"#"make up "A" or "K" these two letters。Makea program to identify these letters, we want to kown the number of these twoletters.
输入
Input contains multiple test cases.
Each test case contains two integer N (10<=N<=100) and M (10<=M<=100) is the graphics of the number of rows and columns。
Next graphics, letter fonts and titles to describe these two letters the same standard fonts, letter size is not necessarily the same, and does not skew, each letter occupies a separate rectangular area itself is connected, do not cross, and other letters or connected.
输出
Each test case output the number of "A" and "K".
样例输入
10 47 _______________________________________________ _____________####________#####______###________ ____________#######_______####_____###_________ ___________###__###________###___#####_________ ________#####____###_______###_###_____________ _________###______###______######______________ ________##############_____###_###_____________ _______###__________###____###___###___________ ______#####________####___###_____###__________ _____###______________###__###_______###_______
样例输出
1 1
思路：DFS判断图形即可。
#include "stdafx.h" //AK吧，少年！ //解题思路： //1.深搜或者广搜遍历 //2.遇到第一个合适的节点检查是“A”还是“K”，A和K的区别是x+1，y-1是不是# //因为是一次遍历就遍历完全所有相连节点，所以只要判断符合条件的节点的特殊位置就可以判断是哪个字母 #include <stdio.h> #include <string.h> const int MAX = 105; int vis[MAX][MAX]; char map[MAX][MAX]; int dir[8][2] = { 0, 1, 0, -1, 1, 0, -1, 0, -1, -1, -1, 1, 1, -1, 1, 1 }; int n, m,ans[2]; void DFS(int x, int y) { vis[x][y] = 1; for (int i = 0; i < 8; i++) { int nx = dir[i][0] + x; int ny = dir[i][1] + y; if (nx >= 0 && ny >= 0 && nx < n && ny < m && !vis[nx][ny] && map[nx][ny] == ‘#‘) { DFS(nx, ny); } } } #include <iostream> #include <queue> using namespace std; struct Point { int x, y; }; queue<Point> q; void BFS(int x, int y) { vis[x][y] = 1; Point p; p.x = x; p.y = y; q.push(p); while (!q.empty()) { Point p = q.front(); q.pop(); int x = p.x; int y = p.y; vis[x][y] = 1; for (int i = 0; i < n; i++) { int nx = dir[i][0] + x; int ny = dir[i][1] + y; if (nx >= 0 && ny >= 0 && nx < n && ny < m && !vis[nx][ny] && map[nx][ny] == ‘#‘) { Point p; p.x = nx; p.y = ny; q.push(p); } } } } int check(int x, int y) { if (map[x + 1][y - 1] == ‘#‘) return 0; else return 1; } int main() { while (scanf("%d %d", &n, &m) != EOF) { memset(vis, 0, sizeof(vis)); memset(ans, 0, sizeof(ans)); for (int i = 0; i < n; i++) { scanf("%s", map[i]); } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (!vis[i][j] && map[i][j] == ‘#‘) { //DFS(i, j); BFS(i, j); ans[check(i, j)] ++; } } } printf("%d %d\n", ans[0], ans[1]); } return 0; }
