浙江16年省赛 I题 People Counting

People Counting
Time Limit: 2 Seconds      Memory Limit: 65536 KB
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.

Input

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.

Output

For each test case, there should be a single line, containing an integer indicating the number of people from the photo.

Sample Input

2
3 3
.O.
/|\
(.)3 4
OOO(
/|\\
()))

Sample Output

1
4

数小人每次删去一个人即可
ACcode：

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)
#define rdl(x) scanf("%I64d,&x);
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define ull unsigned long long
#define maxn 1005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
#define eps 1e-8
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll mul(ll a,ll b,ll p){ll sum=0;for(;b;a=(a+a)%p,b>>=1)if(b&1)sum=(sum+a)%p;return sum;}
inline void Scan(int &x) {
char c;while((c=getchar())<'0' || c>'9');x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
using namespace std;
char dp[maxn][maxn];
int x,y,n,m;
void  gao(){
if(dp[x][y]=='O'){
if(dp[x+1][y]=='|')dp[x+1][y]='.';
if(dp[x+1][y-1]=='/')dp[x+1][y-1]='.';
if(dp[x+1][y+1]=='\\')dp[x+1][y+1]='.';
if(dp[x+2][y-1]=='(')dp[x+2][y-1]='.';
if(dp[x+2][y+1]==')')dp[x+2][y+1]='.';
}
else if(dp[x][y]=='|'){
if(dp[x-1][y]=='O')dp[x-1][y]='.';
if(dp[x][y-1]=='/')dp[x][y-1]='.';
if(dp[x][y+1]=='\\')dp[x][y+1]='.';
if(dp[x+1][y-1]=='(')dp[x+1][y-1]='.';
if(dp[x+1][y+1]==')')dp[x+1][y+1]='.';
}
else if(dp[x][y]=='/'){
if(dp[x-1][y+1]=='O')dp[x-1][y+1]='.';
if(dp[x][y+1]=='|')dp[x][y+1]='.';
if(dp[x][y+2]=='\\')dp[x][y+2]='.';
if(dp[x+1][y]=='(')dp[x+1][y]='.';
if(dp[x+1][y+2]==')')dp[x+1][y+2]='.';
}
else if(dp[x][y]=='\\'){
if(dp[x-1][y-1]=='O')dp[x-1][y-1]='.';
if(dp[x][y-1]=='|')dp[x][y-1]='.';
if(dp[x][y-2]=='/')dp[x][y-2]='.';
if(dp[x+1][y-2]=='(')dp[x+1][y-2]='.';
if(dp[x+1][y]==')')dp[x+1][y]='.';
}
else if(dp[x][y]=='('){
if(dp[x-2][y+1]=='O')dp[x-2][y+1]='.';
if(dp[x-1][y]=='/')dp[x-1][y]='.';
if(dp[x-1][y+1]=='|')dp[x-1][y+1]='.';
if(dp[x-1][y+2]=='\\')dp[x-1][y+2]='.';
if(dp[x][y+2]==')')dp[x][y+2]='.';
}
else if(dp[x][y]==')'){
if(dp[x-2][y-1]=='O')dp[x-2][y-1]='.';
if(dp[x-1][y-2]=='/')dp[x-1][y-2]='.';
if(dp[x-1][y-1]=='|')dp[x-1][y-1]='.';
if(dp[x-1][y]=='\\')dp[x-1][y]='.';
if(dp[x][y-2]==')')dp[x][y-2]='.';
}
dp[x][y]='.';
/*  for(int i=1+5;i<=n+5;++i){
for(int j=1+5;j<=m+5;++j)cout<<dp[i][j];
cout<<'\12';}*/
}
int juge(){
for(int i=1+5;i<=n+5;++i)
for(int j=1+5;j<=m+5;++j)
if(dp[i][j]!='.'){
x=i;
y=j;
gao();
return 1;
}
return 0;
}
void init(){
for(int i=0;i<maxn;++i)for(int j=0;j<maxn;++j)dp[i][j]='.';
}
int main(){
int loop,cnt=1;
rd(loop);
while(loop--){
rd2(n,m);
//   init();
for(int i=1+5;i<=n+5;++i)
for(int j=1+5;j<=m+5;++j)cin>>dp[i][j];
int ans=0;
while(juge())ans++;
printf("%d\n",ans);
}
return 0;
}