Problem 2150 Fire Game
Accept: 3772 Submit: 12868
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Problem DescriptionFat 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：给了你一块草地，‘#‘表示地上有草，火可以蔓延，‘.‘火不可以蔓延，然后有两个吃的没事干的人分别在这个草地上的某一个地方点火，点火的时间不算，之后每一秒火都会蔓延到其上下左右，问可以不可烧完所有的草，可以就输出最少的时间，不可以就输出-1
思路：只能枚举每两个草了，所以题目给的地也不是很大，也就10*10最大了，之后就每两个点一起烧，一开始队列里面要放进去两个点，之后可以烧到的草有可能两把火都会被烧到，所以应该取较早时间被烧到的那个时间，但如果火都烧没了，不能蔓延了却还有草，说明不能烧完所有的草，那就输出-1
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
#define ll long long
int const maxn = 15;
const int mod = 1e9 + 7;
int gcd(int a, int b) {
if (b == 0) return a; return gcd(b, a % b);
}
char maze[maxn][maxn];
int ans;
struct point
{
int x,y;
};
int n,m;
int dis[maxn][maxn];
queue<point>que;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int bfs(int x1,int y1,int x2,int y2)
{
while(!que.empty())
que.pop();
memset(dis,INF,sizeof(dis));
point p1,p2,next;
p1.x=x1;
p1.y=y1;
p2.x=x2;
p2.y=y2;
dis[x1][y1]=0;
dis[x2][y2]=0;
que.push(p1);
que.push(p2);
while(!que.empty())
{
point p=que.front();
que.pop();
for(int i=0;i<4;i++)
{
int nx=p.x+dx[i];
int ny=p.y+dy[i];
if(nx>=0 && nx<n && ny>=0 && ny<m && maze[nx][ny]==‘#‘ && dis[nx][ny]>dis[p.x][p.y]+1)
{
dis[nx][ny]=dis[p.x][p.y]+1;
next.x=nx;
next.y=ny;
que.push(next);
}
}
}
int maxx=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(maze[i][j]==‘#‘)
maxx=max(maxx,dis[i][j]);
return maxx;
}
int main()
{
int t;
scanf("%d",&t);
int ca=1;
while(t--)
{
int ans=INF;
scanf("%d %d",&n,&m);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
cin>>maze[i][j];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(maze[i][j]==‘#‘)
{
for(int ii=0;ii<n;ii++)
{
for(int jj=0;jj<m;jj++)
{
if(maze[ii][jj]==‘#‘)
{
int temp=bfs(i,j,ii,jj);
ans=min(ans,temp);
}
}
}
}
}
}
if(ans==INF)
ans=-1;
printf("Case %d: %d\n",ca++,ans);
}
}
原文地址：https://www.cnblogs.com/smallhester/p/9567845.html