Description
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."Sample Input
7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........Sample Output
13
根据friend和friends,我们可以推知angel有多个朋友
也就是说,这是一个有一个入口多个出口的迷宫
从angel所在位置开始BFS,队列使用优先队列,确保按照距离层数访问节点
(第一次蜜汁RE)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 32 const int maxn = 210; 33 34 int N,M; 35 char Map[maxn][maxn]; 36 int dis[maxn][maxn]; 37 38 struct point { 39 int x,y; 40 int dis; 41 point(int a,int b,int c) { 42 x = a; 43 y = b; 44 dis = c; 45 } 46 bool operator < (const point &rhs)const { 47 return dis > rhs.dis; 48 } 49 }; 50 51 52 const int delta[] = {1,-1,0,0}; 53 54 int BFS(int s1,int s2) { 55 priority_queue<point> Q; 56 bool visited[maxn][maxn]; 57 memset(visited,false,sizeof(visited)); 58 memset(dis,-1,sizeof(dis)); 59 60 Q.push(point(s1,s2,0)); 61 dis[s1][s2] = 0; 62 63 while(!Q.empty()) { 64 int x = Q.top().x; 65 int y = Q.top().y; 66 int dist = Q.top().dis; 67 68 Q.pop(); 69 70 if(visited[x][y] == true) 71 continue; 72 visited[x][y] = true; 73 74 //拓展节点 75 REP(4) { 76 int xx = x + delta[o]; 77 int yy = y + delta[3 - o]; 78 79 if(xx >= 0 && xx < N && yy >= 0 && yy < M && Map[xx][yy] != '#' && visited[xx][yy] == false) { 80 int dd = dist + 1; 81 82 if(Map[xx][yy] == 'x') 83 dd++; 84 85 dis[xx][yy] = ((dis[xx][yy] == -1) ? dd : min(dd,dis[xx][yy])); 86 87 if(Map[xx][yy] == 'r') 88 return dis[xx][yy]; 89 90 Q.push(point(xx,yy,dis[xx][yy])); 91 92 } 93 94 } 95 96 } 97 98 return -1; 99 100 } 101 102 bool Do() { 103 if(scanf("%d%d",&N,&M) == EOF) 104 return false; 105 106 int s1,s2; 107 for(int i = 0; i < N; i++) 108 for(int j = 0; j < M; j++) { 109 scanf(" %c ",&Map[i][j]); 110 if(Map[i][j] == 'a') { 111 Map[i][j] = '.'; 112 s1 = i; 113 s2 = j; 114 } 115 } 116 117 #if debug 118 for(int i = 0; i < N; i++) { 119 for(int j = 0; j < M; j++) 120 printf("%c",Map[i][j]); 121 printf(" "); 122 } 123 #endif 124 125 int ans = BFS(s1,s2); 126 127 if(ans == -1) 128 printf("Poor ANGEL has to stay in the prison all his life. "); 129 else 130 printf("%d ",ans); 131 132 return true; 133 } 134 135 int main() { 136 while(Do()); 137 return 0; 138 }
下面是重新写了一遍的AC代码
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 = 250; 32 int n,m; 33 char Map[maxn][maxn]; 34 35 int s1,s2; 36 37 int dis[maxn][maxn]; 38 39 const int delta[] = {1,-1,0,0}; 40 41 struct point { 42 int x,y; 43 int dis; 44 45 point(int a,int b,int c) { 46 x = a; 47 y = b; 48 dis = c; 49 } 50 51 bool operator < (const point &rhs)const { 52 return dis>rhs.dis; 53 } 54 }; 55 56 int BFS() { 57 priority_queue<point> Q; 58 memset(dis,-1,sizeof(dis)); 59 60 Q.push(point(s1,s2,0)); 61 dis[s1][s2] = 0; 62 while(!Q.empty()) { 63 int x = Q.top().x; 64 int y = Q.top().y; 65 int dist = Q.top().dis; 66 67 Q.pop(); 68 69 REP(4) { 70 int xx = x + delta[o]; 71 int yy = y + delta[3 - o]; 72 int dd = dist + 1; 73 74 if(xx >= 0 && xx < n && yy>=0 && yy < m) { 75 if(Map[xx][yy] != '#' && dis[xx][yy] == -1) { 76 if(Map[xx][yy] == 'x') 77 dd++; 78 if(Map[xx][yy] == 'r') 79 return dd; 80 dis[xx][yy] = dd; 81 Q.push(point(xx,yy,dd)); 82 } 83 } 84 } 85 } 86 return -1; 87 } 88 89 bool Do() { 90 if(scanf("%d%d",&n,&m) == EOF) 91 return false; 92 for(int i = 0;i < n;i++) 93 for(int j = 0;j < m;j++) { 94 scanf(" %c ",&Map[i][j]); 95 if(Map[i][j] == 'a') { 96 s1 = i; 97 s2 = j; 98 } 99 } 100 101 int ans = BFS(); 102 if(ans == -1) 103 printf("Poor ANGEL has to stay in the prison all his life. "); 104 else 105 printf("%d ",ans); 106 107 return true; 108 } 109 110 int main() { 111 while(Do()); 112 return 0; 113 }