In a BG (dinner gathering) for ZJU ICPC team, the coaches wanted to count the number of people present at the BG. They did that by having the waitress take a photo for them. Everyone was in the photo and no one was completely blocked. Each person in the photo has the same posture. After some preprocessing, the photo was converted into a H×W character matrix, with the background represented by ".". Thus a person in this photo is represented by the diagram in the following three lines:
.O. /|\ (.)Given the character matrix, the coaches want you to count the number of people in the photo. Note that if someone is partly blocked in the photo, only part of the above diagram will be presented in the character matrix.
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first contains two integers H, W (1 ≤ H, W ≤ 100) - as described above, followed by H lines, showing the matrix representation of the photo.
For each test case, there should be a single line, containing an integer indicating the number of people from the photo.
题意：统计在一幅照片里统计人的数量 解题思路：每个位置放入一个人，若有人的一部分，则有一个人
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <map> #include <cmath> #include <set> #include <stack> #include <queue> #include <vector> #include <bitset> #include <functional> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; char ch[120][120]; int main() { int t; scanf("%d",&t); while (t--) { int n, m; scanf("%d%d", &n, &m); n += 2; m += 2; for (int i = 2; i < n; i++) scanf("%s", ch[i] + 2); int ans = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (ch[i][j + 1] == 'O' || ch[i + 1][j] == '/' || ch[i + 1][j + 1] == '|' || ch[i + 1][j + 2] == '\\' || ch[i + 2][j] == '(' || ch[i + 2][j + 2] == ')')ans++; } } printf("%d\n", ans); } return 0; }
