题意

给个N＊M的田地，有些地方是草地有些是空地，现在有两个小孩，开始放火。放火只能在草地上进行，并且第一块点火的草地烧掉不需要时间，且每一块着火的草地，每一秒都会点着烧掉周围四块草地。问两个小孩最少要多少时间能把这片田地上的草烧完。每个小孩只有一次放火机会。

分析

N ， M 不会很大，直接枚举两个人点的草地位置，分别bfs，在所有解里取最小值即可。

代码1

/* When all else is lost the future still remains. */ /* You can be the greatest */ #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++) #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--) #define fi first #define se second #define mk(X,Y) make_pair((X),(Y)) #define inf 0x3f3f3f3f #define clr(X,Y) memset(X,Y,sizeof(X)) #define pb push_back //head #include <iostream> #include <stdio.h> #include <queue> #include <algorithm> #include <string> #include <map> #include <string.h> using namespace std; #define maxN 15 #define maxM 15 int dx[] = {-1,0,1,0}; int dy[] = {0,-1,0,1}; int val[2][maxN][maxM]; int mark[2][maxN][maxM]; bool mp[maxN][maxM]; int n , m; int check_val(int,int); int check_val(int n , int m){ int ans = 0; rep(i,1,n+1) rep(j,1,m+1){ if(mp[i][j]){ ans = max(ans,min(val[1][i][j],val[0][i][j])); } } return ans; } int dfs(int x , int y , int cnt){ queue<pair<int,int> > Q; val[cnt][x][y] = 0; mark[cnt][x][y] = 1; Q.push(mk(x,y)); while(!Q.empty()){ int nx = Q.front().fi; int ny = Q.front().se; Q.pop(); rep(i,0,4){ int tx = nx + dx[i]; int ty = ny + dy[i]; if(!mp[tx][ty]) continue; if(mark[cnt][tx][ty]) continue; val[cnt][tx][ty] = val[cnt][nx][ny] + 1; mark[cnt][tx][ty] = 1; Q.push(mk(tx,ty)); } } if(cnt == 1) return check_val(n,m); int ans = inf; rep(i,x,n+1){ int t = (i == x) ? (y) : 1; rep(j,t,m+1){ if(mp[i][j]){ rep(k,1,n+1) rep(l,1,m+1) mark[1][k][l] = 0; rep(k,1,n+1) rep(l,1,m+1) val[1][k][l] = inf; ans = min(ans,dfs(i,j,1)); } } } return ans; } int main(){ int T; while(~scanf("%d",&T)) rep(ca,1,T+1){ clr(mp,0); scanf("%d %d\n",&n,&m); char t; rep(i,1,n+1) { rep(j,1,m+1){ scanf("%c",&t); mp[i][j] = (t == '.') ? 0 : 1; } getchar(); } int ans = inf; rep(i,1,n+1) rep(j,1,m+1){ if(mp[i][j]){ clr(val,inf); clr(mark,0); ans = min(ans,dfs(i,j,0)); } } printf("Case %d: %d\n",ca,(ans>=inf)?-1:ans); } return 0; }

代码2

 1 /* When all else is lost the future still remains. */  2 /* You can be the greatest */  3 #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++)  4 #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--)  5 #define fi first  6 #define se second  7 #define mk(X,Y) make_pair((X),(Y))  8 #define inf 0x3f3f3f3f  9 #define clr(X,Y) memset(X,Y,sizeof(X)) 10 #define pb push_back 11 //head 12 #include <iostream> 13 #include <stdio.h> 14 #include <queue> 15 #include <algorithm> 16 #include <string> 17 #include <map> 18 #include <string.h> 19 using namespace std; 20 #define maxN 20 21 #define maxM 20 22 int dx[] = {-1,0,1,0}; 23 int dy[] = {0,-1,0,1}; 24 int n , m; 25 bool mp[maxN][maxM]; 26 int val[2][maxN][maxM]; 27 bool mark[2][maxN][maxM]; 28 void init(int cnt){ 29 rep(i,1,n+1) rep(j,1,m+1){ 30 val[cnt][i][j] = inf; 31 mark[cnt][i][j] = 0; 32  } 33 return ; 34 } 35 int check_val(){ 36 int ans = 0; 37 rep(i,1,n+1) rep(j,1,m+1){ 38 if(mp[i][j]) ans = max(ans,min(val[0][i][j],val[1][i][j])); 39  } 40 return ans; 41 } 42 void bfs(int x , int y , int cnt){ 43 queue<pair<int,int> > Q; 44  Q.push(mk(x,y)); 45 mark[cnt][x][y] = 1; 46 val[cnt][x][y] = 0; 47 while(!Q.empty()){ 48 int nx = Q.front().fi; 49 int ny = Q.front().se; 50  Q.pop(); 51 rep(i,0,4){ 52 int tx = nx + dx[i]; 53 int ty = ny + dy[i]; 54 if(!mp[tx][ty]) continue; 55 if(mark[cnt][tx][ty]) continue; 56 val[cnt][tx][ty] = val[cnt][nx][ny] + 1; 57 mark[cnt][tx][ty] = 1; 58  Q.push(mk(tx,ty)); 59  } 60  } 61 } 62 int main(){ 63 64 int T; 65 scanf("%d",&T); 66 rep(ca,1,T+1){ 67 int ans = inf; 68 clr(mp,0); 69 scanf("%d %d\n",&n,&m); 70 char t; 71 rep(i,1,n+1) { 72 rep(j,1,m+1){ 73 scanf("%c",&t); 74 mp[i][j] = (t == '.') ? 0 : 1; 75  } 76  getchar(); 77  } 78 rep(rx,1,n+1) rep(ry,1,m+1){ 79 if(!mp[rx][ry]) continue; 80 init(0); init(1); 81 bfs(rx,ry,0); 82 rep(ax,rx,n+1){ 83 int t = (ax == rx) ? ry : 1; 84 rep(ay,t,m+1){ 85 if(!mp[ax][ay]) continue; 86 init(1); 87 bfs(ax,ay,1); 88 ans = min(ans,check_val()); 89  } 90  } 91  } 92 printf("Case %d: %d\n",ca,(ans>=inf)?-1:ans); 93 94  } 95 return 0; 96 }