Accept: 693 Submit: 2657 Time Limit: 1000 mSec Memory Limit : 32768 KB
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
Sample Output
题目描述：
给定一张图。图中"#"代表草，'.'代表地，可能有几个草地，给你两把火，问烧完整个图中的草的最短时间。
由于n<=10,所以设立一个结构体，里面的属性有位置，和被烧的时间，然后枚举两把火的位置，开始烧，
用cnt计数烧过的草，然后将枚举的两个位置放进队列，如果两个位置重合的话。cnt=1，否则cnt=2；
其实我们可以认为i这两把火是有一把火烧起来的，那么之后就跟裸的宽搜是一样的,（其实跟锁屏样式的两种判断转换成一种差不多）time，每层+1.#include <iostream> #include <cstdio> #include <cmath> #include <queue> #include <cstring> #define inf 0x3f3f3f3f using namespace std; char MAP[20][20]; int visit[20][20]; int dir[4][2]={-1,0,1,0,0,1,0,-1}; int n,m,sum,cnt; struct node { int x, y,Time; }; int main() { int t; scanf("%d",&t); int _case=0; while(t--) { queue< node > Q; sum=0; scanf("%d%d",&n,&m); for(int i=0;i<n;i++) { getchar(); for(int j=0;j<m;j++) { scanf("%c",&MAP[i][j]); if(MAP[i][j]=='#') { ++sum; } } } int answer=inf; int cntll=0; for(int i=0;i<n;i++) for(int j=0;j<m;j++) { if(MAP[i][j] == '#') { for(int xx=i;xx<n;xx++) for(int yy=0;yy<m;yy++) { if(xx==i && yy<j) continue; memset(visit,0,sizeof(visit)); if(MAP[xx][yy]=='#') { cnt=2; if(i==xx && j==yy) { cnt=1; } visit[i][j]=1; visit[xx][yy]=1; node NODE1,NODE2,NODE; NODE1.x=i; NODE1.y=j;NODE1.Time=0; NODE2.x=xx; NODE2.y=yy;NODE2.Time=0; Q.push(NODE1); Q.push(NODE2); while(!Q.empty()) { NODE=Q.front(); Q.pop(); node point; for(int k=0;k<4;k++) { point.x=NODE.x+dir[k][0]; point.y=NODE.y+dir[k][1]; if(point.x>=0 && point.x<n && point.y>=0 && point.y<m) if(visit[point.x][point.y]==0 && MAP[point.x][point.y]=='#') { visit[point.x][point.y]=1; point.Time=NODE.Time+1; Q.push(point); cnt=cnt+1; } } } if(cnt==sum) { answer=min(answer,NODE.Time); } } } } } if(answer!=inf) printf("Case %d: %d ",++_case,answer); else printf("Case %d: -1 ",++_case); } return 0; }