FZUOJ 2150 Fire Game

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2150

题目大意：小明小红要在一块网格状土地上玩火。土地上有的地方是草块，有的地方是空地。草地会被相邻格子的火引燃，土地不会。小明小红想要让所有有草的格子都燃烧，并且它俩各自点着了一块草地（可能一样）。问是否能够达到他们的目的？如果能，最少需要多长时间才能烧完全部的草地。

解题思路：显然有三个以上互相不连通的草地块那么就无法达到目的了，如果有两个，那必须一人点一个草地块，如果有一个就任选两个格子点燃，即：

DFS求连通块个数，若两个-对于每个块，枚举每个点，BFS求最少时间然后求两个最小时间的最大值；若一个-枚举这个快内任意两个点（可以一样），然后两个点同时为起点bfs即可，最后求最小值即可。

代码：

  1 const int inf = 0x3f3f3f3f;
  2 const int iadd[] = {0, 1, 0, -1}, jadd[] = {1, 0, -1, 0};
  3 const int maxn = 15;
  4 char maze[maxn][maxn];
  5 int n, m;
  6 int bt[maxn][maxn], cnt = 0;
  7 int anss[3];
  8 short vis[maxn][maxn];
  9
 10 void dfs(int x, int y){
 11     vis[x][y] = 1; bt[x][y] = cnt + 1;
 12     for(int i = 0; i < 4; i++){
 13         int vi = x, vj = y;
 14         vi += iadd[i]; vj += jadd[i];
 15         if(vi < 0 || vj < 0 || vi >= n || vj >= m) continue;
 16         if(maze[vi][vj] == '.' || vis[vi][vj]) continue;
 17         dfs(vi, vj);
 18     }
 19 }
 20
 21 struct node{
 22     int i, j, t;
 23 };
 24 int bfs(int i, int j){
 25     queue<node> q;
 26     memset(vis, 0, sizeof(vis));
 27     vis[i][j] = 1;
 28     node u; u.i = i; u.j = j; u.t = 0;
 29     q.push(u);
 30     int ans = 0;
 31     while(!q.empty()){
 32         u = q.front(); q.pop();
 33         ans = u.t; u.t++;
 34         for(int i = 0; i < 4; i++){
 35             node v = u;
 36             v.i += iadd[i]; v.j += jadd[i];
 37             if(v.i < 0 || v.j < 0 || v.i >= n || v.j >= m) continue;
 38             if(maze[v.i][v.j] == '.' || vis[v.i][v.j]) continue;
 39             vis[v.i][v.j] = 1;
 40             q.push(v);
 41         }
 42     }
 43     anss[bt[i][j]] = min(anss[bt[i][j]], ans);
 44 }
 45
 46 int bfs(int i1, int j1, int i2, int j2){
 47     memset(vis, 0, sizeof(vis));
 48     queue<node> q;
 49     vis[i1][j1] = vis[i2][j2] = 1;
 50     node u1; u1.i = i1; u1.j = j1; u1.t = 0;
 51     node u2; u2.i = i2; u2.j = j2; u2.t = 0;
 52     q.push(u1); q.push(u2);
 53     int ans = 0;
 54     while(!q.empty()){
 55         node u = q.front(); q.pop();
 56         ans = max(ans, u.t);
 57         u.t++;
 58         for(int i = 0; i < 4; i++){
 59             node v = u;
 60             v.i += iadd[i]; v.j += jadd[i];
 61             if(v.i < 0 || v.j < 0 || v.i >= n || v.j >= m) continue;
 62             if(maze[v.i][v.j] == '.' || vis[v.i][v.j]) continue;
 63             q.push(v);
 64             vis[v.i][v.j] = 1;
 65         }
 66     }
 67     return ans;
 68 }
 69
 70 int main(){
 71     int T;
 72     scanf("%d", &T);
 73     for(int t = 1; t <= T; t++){
 74         memset(bt, 0, sizeof(bt));
 75         cnt = 0;
 76         scanf("%d %d", &n, &m);
 77         for(int i = 0; i < n; i++){
 78             scanf("%s", maze[i]);
 79         }
 80         printf("Case %d: ", t);
 81         memset(vis, 0, sizeof(vis));
 82         for(int i = 0; i < n; i++){
 83             for(int j = 0; j < m; j++){
 84                 if(maze[i][j] == '#' && !vis[i][j]){
 85                     dfs(i, j);
 86                     cnt++;
 87                 }
 88             }
 89         }
 90         if(cnt == 0) {
 91             puts("0");
 92             continue;
 93         }
 94         else if(cnt >= 3){
 95             puts("-1");
 96             continue;
 97         }
 98         else if(cnt == 1){
 99             vector<PII> va;
100             for(int i = 0; i < n; i++)
101                 for(int j = 0; j < m; j++)
102                     if(maze[i][j] == '#')
103                         va.push_back(PII(i, j));
104             int ans = inf;
105             for(int i = 0; i < va.size(); i++){
106                 for(int j = i; j < va.size(); j++){
107                     int tmp = bfs(va[i].first, va[i].second, va[j].first, va[j].second);
108                     ans = min(tmp, ans);
109                 }
110             }
111             printf("%d\n", ans);
112             continue;
113         }
114         memset(vis, 0, sizeof(vis));
115         memset(anss, 0x3f, sizeof(anss));
116         for(int i = 0; i < n; i++){
117             for(int j = 0; j < m; j++){
118                 if(maze[i][j] == '#') bfs(i, j);
119             }
120         }
121         if(anss[1] >= inf) anss[1] = -1;
122         if(anss[2] >= inf) anss[2] = -1;
123         printf("%d\n", max(anss[2], anss[1]));
124     }
125 }

题目：

Problem 2150 Fire Game

Accept: 2635 Submit: 9094
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

4 3 3 .#. ### .#. 3 3 .#. #.# .#. 3 3 ... #.# ... 3 3 ### ..# #.#

Sample Output

Case 1: 1 Case 2: -1 Case 3: 0 Case 4: 2