http://acm.fzu.edu.cn/problem.php?pid=2150
Problem Description
Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)
You can assume that the grass in the board would never burn out and the empty grid would never get fire.
Note that the two grids they choose can be the same.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains two integers N and M indicate the size of the board. Then goes N line, each line with M character shows the board. “#” Indicates the grass. You can assume that there is at least one grid which is consisting of grass in the board.
1 <= T <=100, 1 <= n <=10, 1 <= m <=10
Output
For each case, output the case number first, if they can play the MORE special (hentai) game (fire all the grass), output the minimal time they need to wait after they set fire, otherwise just output -1. See the sample input and output for more details.
Sample Input
4 3 3 .#. ### .#. 3 3 .#. #.# .#. 3 3 ... #.# ... 3 3 ### ..# #.#
Sample Output
Case 1: 1 Case 2: -1 Case 3: 0 Case 4: 2
题意分析：
两个人在玩游戏，在一块草地上，有的地方是木板，木板不可被烧，两个人各选取一个地方开始烧火， 每一秒钟火会向四周蔓延，求如果两个人的起始烧火位置都是最佳的，最少多少秒可以烧光所有草地。
解题思路：
思路1：遍历所有的点，如果两点都是草地，那么对两点进行BFS，更新最短时间。
思路2：先把草地的点存下来，再遍历两点BFS。略快于思路1。
思路1代码：
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define N 12
using namespace std;
char str[N][N];
bool book[N][N];
int n, m, ans;
struct edge{
int x;
int y;
int temp;
edge(int x, int y, int temp) : x(x), y(y), temp(temp) {}
};
int bfs(int s1, int s2, int s3, int s4)
{
memset(book, false, sizeof(book));
int nex[4][2]={0,1, 0,-1, 1,0, -1,0};
book[s1][s2]=book[s3][s4]=true;
queue<edge>q;
q.push(edge(s1, s2, 0));
q.push(edge(s3, s4, 0));
int sum=0;
while(!q.empty())
{
edge u=q.front();
q.pop();
sum=max(sum, u.temp);
if(sum>ans)
return -1;
for(int i=0; i<4; i++)
{
int tx=u.x+nex[i][0];
int ty=u.y+nex[i][1];
if(tx>=n || ty>=m || tx<0 || ty<0 || book[tx][ty]==true)
continue;
if(str[tx][ty]=='#')
{
q.push(edge(tx, ty, u.temp+1));
book[tx][ty]=true;
}
}
}
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
if(book[i][j]==false && str[i][j]=='#')
return -1;
return sum;
}
int main()
{
int t=0, T;
scanf("%d", &T);
while(T--)
{
ans=0x3f3f3f3f;
scanf("%d%d", &n, &m);
for(int i=0; i<n; i++)
scanf("%s", str[i]);
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
for(int p=0; p<n; p++)
for(int q=0; q<m; q++)
{
if(str[i][j]=='#' && str[p][q]=='#')
{
int s=bfs(i, j, p, q);
if(s!=-1 && s<ans)
{
ans=s;
//printf("%d %d %d %d
", i, j, p, q);
}
}
}
if(ans!=0x3f3f3f3f)
printf("Case %d: %d
", ++t, ans);
else
printf("Case %d: -1
", ++t);
}
return 0;
} 思路2代码：
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define N 20
using namespace std;
char str[N][N];
bool book[N][N];
int n, m, ans;
struct edge{
int x;
int y;
int temp;
edge(){}
edge(int x, int y, int temp) : x(x), y(y), temp(temp) {}
}a[N*N];
int bfs(int s1, int s2, int s3, int s4)
{
memset(book, false, sizeof(book));
int nex[4][2]={0,1, 0,-1, 1,0, -1,0};
book[s1][s2]=book[s3][s4]=true;
queue<edge>q;
q.push(edge(s1, s2, 0));
q.push(edge(s3, s4, 0));
int sum=0;
while(!q.empty())
{
edge u=q.front();
q.pop();
sum=max(sum, u.temp);
if(sum>ans)
return -1;
for(int i=0; i<4; i++)
{
int tx=u.x+nex[i][0];
int ty=u.y+nex[i][1];
if(tx>=n || ty>=m || tx<0 || ty<0 || book[tx][ty]==true)
continue;
if(str[tx][ty]=='#')
{
q.push(edge(tx, ty, u.temp+1));
book[tx][ty]=true;
}
}
}
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
if(book[i][j]==false && str[i][j]=='#')
return -1;
return sum;
}
int main()
{
int t=0, T;
scanf("%d", &T);
while(T--)
{
ans=0x3f3f3f3f;
scanf("%d%d", &n, &m);
for(int i=0; i<n; i++)
scanf("%s", str[i]);
int len=0;
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
if(str[i][j]=='#')
{
a[len].x=i;
a[len].y=j;
len++;
}
for(int i=0; i<len; i++)
for(int j=i; j<len; j++)
{
int s=bfs(a[i].x, a[i].y, a[j].x, a[j].y);
if(s!=-1 && s<ans)
{
ans=s;
}
}
if(ans!=0x3f3f3f3f)
printf("Case %d: %d
", ++t, ans);
else
printf("Case %d: -1
", ++t);
}
return 0;
}