HDU-5706-GirlCat【BFS】【2016CCPC女生专场】

HDU-5706-GirlCat

Problem Description
As a cute girl, Kotori likes playing Hide and Seek'' with cats particularly.
Under the influence of Kotori, many girls and cats are playingHide and Seek” together.
Koroti shots a photo. The size of this photo is n×m, each pixel of the photo is a character of the lowercase(from a' toz’).
Kotori wants to know how many girls and how many cats are there in the photo.

We define a girl as – we choose a point as the start, passing by 4 different connected points continuously, and the four characters are exactly girl'' in the order.
We define two girls are different if there is at least a point of the two girls are different.
We define a cat as -- we choose a point as the start, passing by 3 different connected points continuously, and the three characters are exactlycat” in the order.
We define two cats are different if there is at least a point of the two cats are different.

Two points are regarded to be connected if and only if they share a common edge.

Input
The first line is an integer T which represents the case number.

As for each case, the first line are two integers n and m, which are the height and the width of the photo.
Then there are n lines followed, and there are m characters of each line, which are the the details of the photo.

It is guaranteed that:
T is about 50.
1≤n≤1000.
1≤m≤1000.
∑(n×m)≤2×106.

Output
As for each case, you need to output a single line.
There should be 2 integers in the line with a blank between them representing the number of girls and cats respectively.

Please make sure that there is no extra blank.

Sample Input
3
1 4
girl
2 3
oto
cat
3 4
girl
hrlt
hlca

Sample Output
1 0
0 2
4 1

题目链接：HDU-5706

题目大意：如图所示，分别找出图中完全等于‘girl’和‘cat’的有几种。

注意；
1.必须完全等于‘girl’和‘cat’
2.顺序不能反

题目思路：BFS，以girl为例，找到g进行bfs,旁边等于i的放入队列，以此，看看最后队列里面有几个l，就是答案

//这道题，比赛的时候看错题目意思，以为只要是girl或者irl,单独i也可以。算是一个女孩。代码写的很复杂到最后明白已经晚了，尽管A了，可是没时间写别的题目，感觉好可惜。。

以下是代码：

//
// 5706.cpp
// 2016-CCPC-GIRL
//
// Created by pro on 16/7/3.
// Copyright (c) 2016年 loy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>
using namespace std;
int n,m;
string s[1005];
int vis[1005][1005];
map <char,char> mp;
int dx[5] = {1,-1,0,0};
int dy[5] = {0,0,1,-1};
struct node
{
    char ch;
    int x,y;
};
int ans_girl,ans_cat;
void bfs(int r,int c,char ch)
{
    queue <node> que;

    //头结点
    node zero;
    zero.x = r; zero.y = c; zero.ch = ch;
    que.push(zero);
    while(!que.empty())
    {
        node front = que.front();
        que.pop();
        vis[front.x][front.y] = 1;
        if (front.ch == 'l')
        {
            ans_girl++;
            continue;
        }
        if (front.ch == 't')
        {
            ans_cat++;
            continue;
        }

        for (int i = 0; i < 4; i++)
        {
            int x = dx[i] + front.x;
            int y = dy[i] + front.y;
            if (x < 0 || x >= n || y < 0 || y >= m) continue;  //超出范围
            if (s[x][y] != mp[front.ch]) continue;  //不是接下来的
            node tmp;
            tmp.x = x; tmp.y = y; tmp.ch = s[x][y];
            que.push(tmp);
        }
    }
}
int main()
{
    int t;
    cin >> t;
    mp['g'] = 'i'; mp['i'] = 'r'; mp['r'] = 'l';
    mp['c'] = 'a'; mp['a'] = 't';
    while(t--)
    {
        //初始化
        memset(vis,0,sizeof(vis));
        ans_girl = 0;
        ans_cat = 0;

        cin >> n >> m;
        for (int i = 0; i < n; i++) cin >> s[i];
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (!vis[i][j] && s[i][j] == 'g' || s[i][j] == 'c')
                {
                    bfs(i,j,s[i][j]);
                }
            }
        }
        cout << ans_girl << " " << ans_cat << endl;
    }
    return 0;
}