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
InputThe 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
43 3.#.###.#.3 3.#.#.#.#.3 3...#.#...3 3###..##.#
Sample Output
Case 1: 1Case 2: -1Case 3: 0Case 4: 2
题意：两个小孩分别在两个地方放火，问是否能烧光草地，若能求最少时间。
由于n,m<=10，考虑暴力枚举搜索（BFS)求最少时间。。附上（10000MS的代码）
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
int dr[][2]={{0,1},{0,-1},{-1,0},{1,0}};
//typedef pair<int,int>pill;
struct point {
int x, y;
point() {}
point(int a, int b) { x = a; y = b; }
};
char mp[20][20];
int vis[20][20];
int d[20][20];
int n,m,t,ans;
int bfs(int x1,int y1,int x2,int y2)
{
CLEAR(d,INF);
queue<point>q;
point st;
d[x1][y1]=d[x2][y2]=0;
q.push(point(x1,y1));
q.push(point(x2,y2));
while(!q.empty())
{
st=q.front();
q.pop();
REP(i,4)
{
int xx=st.x+dr[i][0];
int yy=st.y+dr[i][1];
if(xx<0||xx>=n||yy<0||yy>=m) continue;
if(mp[xx][yy]!='#') continue;
if(d[xx][yy]>d[st.x][st.y]+1)
{
d[xx][yy]=d[st.x][st.y]+1;
q.push(point(xx,yy));
}
}
}
int sum=0;
REP(i,n)
{
REP(j,m)
{
if(mp[i][j]=='#')
sum=max(sum,d[i][j]);
}
}
return sum;
}
void solve()
{
ans=INF;
REP(i,n)
{
REP(j,m)
{
if(mp[i][j]=='#')
{
REP(k,n)
{
REP(l,m)
{
if(k==i&&l<=j) continue;//改了就超时
if(mp[k][l]=='#')
ans=min(ans,bfs(i,j,k,l));
}
}
}
}
}
// cout<<"2333 "<<ans<<endl;
printf("%d\n",ans==INF?-1:ans);
}
int main()
{
// std::ios::sync_with_stdio(false);
scanf("%d",&t);
int cas=1;
while(t--)
{
scanf("%d%d",&n,&m);
getchar();
int num=0;
REP(i,n)
{
gets(mp[i]);
REP(j,m)
{
if(mp[i][j]=='#')
num++;
}
}
printf("Case %d: ",cas++);
if(num<=2)
{
cout<<0<<endl;
continue;
}
solve();
}
return 0;
}
附上我的超时的：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
int dr[][2]={{-1,0},{1,0},{0,-1},{0,1}};
typedef pair<int,int>pill;
char mp[15][15];
int vis[15][15];
int d[15][15];
int n,m,t,ans;
int bfs(int x1,int y1,int x2,int y2)
{
CLEAR(d,INF);
queue<pill>q;
pill st;
d[x1][y1]=d[x2][y2]=0;
q.push(make_pair(x1,y1));
q.push(make_pair(x2,y2));
while(!q.empty())
{
st=q.front();
q.pop();
REP(i,4)
{
int xx=st.first+dr[i][0];
int yy=st.second+dr[i][1];
if(xx<0&&xx>=n&&yy<0&&yy>=m) continue;
if(mp[xx][yy]!='#') continue;
if(d[xx][yy]>d[st.first][st.second]+1)
{
d[xx][yy]=d[st.first][st.second]+1;
q.push(make_pair(xx,yy));
}
}
}
int sum=0;
REP(i,n)
{
REP(j,m)
{
if(mp[i][j]=='#')
sum=max(sum,d[i][j]);
}
}
return sum;
}
void solve()
{
ans=INF;
REP(i,n)
{
REP(j,m)
{
if(mp[i][j]=='#')
{
REP(k,n)
{
REP(l,m)
{
if(k<=i&&l<=j) continue;
if(mp[k][l]=='#')
ans=min(ans,bfs(i,j,k,l));
}
}
}
}
}
// cout<<"2333 "<<ans<<endl;
printf("%d\n",ans==INF?-1:ans);
}
int main()
{
std::ios::sync_with_stdio(false);
cin>>t;
int cas=1;
while(t--)
{
cin>>n>>m;
int num=0;
REP(i,n)
{
REP(j,m)
{
cin>>mp[i][j];
if(mp[i][j]=='#')
num++;
}
}
printf("Case %d: ",cas++);
if(num<=2)
{
cout<<0<<endl;
continue;
}
solve();
}
return 0;
}