HDU - 5706 GirlCat（dfs）
GirlCat
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1348 Accepted Submission(s): 829
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 playing ``Hide 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' to `z').
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 exactly ``cat'' 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.
Under the influence of Kotori, many girls and cats are playing ``Hide 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' to `z').
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 exactly ``cat'' 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.
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.
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
Source
Recommend
liuyiding
就是让你在这个字符数组里找到有几个girl和cat~~用dfs一搜索就行了；不过要注意用getchar接收回车键；
#include<iostream>
#include<cstdio>
using namespace std;
char m[1005][1005];
int ans1 = 0;
int ans2 = 0;
int a, b;
int fx[4][2] = { 1,0,0,1,-1,0,0,-1 };
bool check1(int x, int y, int cnt)
{
char w;
if (cnt == 1)
{
w = 'i';
}
else
{
w = cnt == 2 ? 'r' : 'l';
}
// cout << m[x][y] << " " << w <<" "<<x<<" "<<y<< endl;
if (m[x][y] == w)
{
return 1;
}
return 0;
}
bool check2(int x, int y, int cnt)
{
char w;
w = cnt == 1 ? 'a' : 't';
if (m[x][y] == w)
{
return 1;
}
return 0;
}
void bfs1(int x, int y, int cnt)
{
if (cnt == 3)
{
ans1++;
return;
}
for (int s = 0; s < 4; s++)
{
int xx = x + fx[s][0];
int yy = y + fx[s][1];
if (xx == 0 || yy == 0 || xx > a || yy > b)
{
continue;
}
if (check1(xx, yy, cnt + 1))
{
// cout << "1" << endl;
bfs1(xx, yy, cnt + 1);
}
}
}
void bfs2(int x, int y, int cnt)
{
if (cnt == 2)
{
ans2++;
return;
}
for (int s = 0; s < 4; s++)
{
int xx = x + fx[s][0];
int yy = y + fx[s][1];
if (xx == 0 || yy == 0 || xx > a || yy > b)
{
continue;
}
if (check2(xx, yy, cnt + 1))
{
bfs2(xx, yy, cnt + 1);
}
}
}
int main()
{
int n;
cin >> n;
while (n--)
{
ans1 = 0;
ans2 = 0;
scanf("%d%d", &a, &b);
for (int s = 1; s <= a; s++)
{
getchar();
for (int e = 1; e <= b; e++)
{
// cout << s << " " << e << endl;
scanf("%c", &m[s][e]);
}
}
for (int s = 1; s <= a; s++)
{
for (int e = 1; e <= b; e++)
{
if (m[s][e] == 'g')
{
bfs1(s, e, 0);
}
if (m[s][e] == 'c')
{
bfs2(s, e, 0);
}
}
}
cout << ans1 << " " << ans2 << endl;
}
return 0;
}
，这种没能力进去了也白搭，做好13级被输出的准备吧
投递旷视等公司10个岗位 >