Description
Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.Input
The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCFOutput
For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.Sample Input
4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#Sample Output
66 88 66
有两个起点的BFS，每个KFS是一个出口，分别计算出每个出口距离两个入口的距离，计算其最小值
要特别注意：KFC有可能不能到达
1 /* 2 By:OhYee 3 Github:OhYee 4 Email:oyohyee@oyohyee.com 5 Blog:http://www.cnblogs.com/ohyee/ 6 7 かしこいかわいい？ 8 エリーチカ！ 9 要写出来Хорошо的代码哦~ 10 */ 11 12 #include <cstdio> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <string> 17 #include <iostream> 18 #include <vector> 19 #include <list> 20 #include <queue> 21 #include <stack> 22 #include <map> 23 using namespace std; 24 25 //DEBUG MODE 26 #define debug 0 27 28 //循环 29 #define REP(n) for(int o=0;o<n;o++) 30 31 const int maxn = 205; 32 int n,m; 33 char Map[maxn][maxn]; 34 const int delta[] = {1,-1,0,0}; 35 36 struct point { 37 int x,y; 38 point() { 39 x = y = -1; 40 } 41 point(int a,int b) { 42 x = a; 43 y = b; 44 } 45 }; 46 int num; 47 48 49 int BFS(point s,int (&dis)[maxn][maxn]) { 50 memset(dis,-1,sizeof(dis)); 51 queue<point> Q; 52 53 Q.push(s); 54 dis[s.x][s.y] = 0; 55 56 while (!Q.empty()) { 57 int x = Q.front().x; 58 int y = Q.front().y; 59 Q.pop(); 60 61 REP(4) { 62 int xx = x + delta[o]; 63 int yy = y + delta[3 - o]; 64 65 if (xx < 0 || xx >= n || yy < 0 || yy >= m) 66 continue; 67 if (Map[xx][yy] == '#') 68 continue; 69 if (dis[xx][yy] == -1) { 70 dis[xx][yy] = dis[x][y] + 1; 71 Q.push(point(xx,yy)); 72 } 73 } 74 } 75 return -1; 76 } 77 78 bool Do() { 79 if (scanf("%d%d",&n,&m) == EOF) 80 return false; 81 point s1,s2; 82 num = 0; 83 point v[maxn*maxn]; 84 for (int i = 0;i < n;i++) 85 for (int j = 0;j < m;j++) { 86 scanf(" %c",&Map[i][j]); 87 if (Map[i][j] == 'Y') 88 s1 = point(i,j); 89 if (Map[i][j] == 'M') 90 s2 = point(i,j); 91 if (Map[i][j] == '@') 92 v[num++] = point(i,j); 93 } 94 95 int dis1[maxn][maxn]; 96 int dis2[maxn][maxn]; 97 98 BFS(s1,dis1); 99 BFS(s2,dis2); 100 101 int Min = 100000; 102 for (int i = 0;i < num;i++) 103 if(dis1[v[i].x][v[i].y]!=-1&& dis2[v[i].x][v[i].y]!=-1) 104 Min = min(Min,dis1[v[i].x][v[i].y]+ dis2[v[i].x][v[i].y]); 105 106 printf("%d ",Min * 11); 107 108 return true; 109 } 110 111 int main() { 112 while (Do()); 113 return 0; 114 }