题目连接
题目大意就是输入M,N。然后输入map，如果遇到.那么就可一走，耗时为一，如果不为‘x’就是遇到守卫，需要杀死，额外耗时一，'#'是墙。
代码：
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1 #include <stdio.h> 2 #include <string.h> 3 struct node 4 { 5 int x,y; 6 int step; 7 }q[100000]; 8 char map[250][250]; 9 int vis[250][250]; 10 int to[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};//方向 11 int f,r,m,n,leap,flag; 12 13 void sort()//优先队列的排序 14 { 15 int i; 16 for(i = r;i > f;i--) 17 { 18 struct node temp; 19 if(q[i].step < q[i-1].step) 20 { 21 temp = q[i]; 22 q[i] = q[i-1]; 23 q[i-1] = temp; 24 } 25 else 26 break; 27 } 28 } 29 void bfs(int px,int py)//bfs 30 { 31 memset(vis,0,sizeof(vis)); 32 int i; 33 f = r = 0; 34 q[r].x = px; 35 q[r].y = py; 36 q[r].step = 0; 37 r++; 38 39 while(f < r) 40 { 41 struct node now,temp; 42 now = q[f++]; 43 for(i = 0;i < 4;i++) 44 { 45 temp.x = now.x + to[i][0]; 46 temp.y = now.y + to[i][1]; 47 temp.step = now.step+1; 48 if(temp.x >= 0 && temp.x < m && temp.y >= 0 && temp.y < n)//查边界 49 { 50 if(map[temp.x][temp.y] != '#' && !vis[temp.x][temp.y]) 51 { 52 vis[temp.x][temp.y] = 1; 53 if(map[temp.x][temp.y] == '.') 54 { 55 q[r] = temp; 56 sort(); 57 } 58 else if(map[temp.x][temp.y] == 'x') 59 { 60 temp.step++; 61 q[r] = temp; 62 sort(); 63 } 64 else if(map[temp.x][temp.y] == 'r') 65 { 66 q[r] = temp; 67 sort(); 68 leap = 1; 69 break; 70 } 71 r++; 72 } 73 } 74 } 75 if(leap) 76 break; 77 } 78 return ; 79 } 80 int main() 81 { 82 int i,j; 83 84 while(~scanf("%d %d",&m,&n)) 85 { 86 leap = flag = 0; 87 memset(map,0,sizeof(map)); 88 89 for(i = 0;i < m;i++) 90 scanf("%s",map[i]); 91 92 for(i = 0;i < m;i++) 93 { 94 for(j = 0;j < n;j++) 95 { 96 if(map[i][j] == 'a')//找到入口 97 { 98 flag = 1; 99 bfs(i,j); 100 break; 101 } 102 } 103 if(flag) 104 break; 105 106 } 107 if(leap) 108 printf("%d\n",q[r].step); 109 else 110 puts("Poor ANGEL has to stay in the prison all his life."); 111 } 112 return 0; 113 }
1 #include <stdio.h> 2 #include <iostream> 3 #include <string.h> 4 5 using namespace std; 6 7 int vis[10005]; 8 int prim[10005],leap; 9 struct node 10 { 11 int n[5],step; 12 int num; 13 }q[4*10005]; 14 void make_prim() 15 { 16 memset(prim,0,sizeof(prim)); 17 int flag = 1,i,j; 18 for(i = 1001;i <= 9999;i += 2) 19 { 20 flag = 1; 21 for(j = 2;j <= i/2;j++) 22 { 23 if(i%j == 0) 24 { 25 flag = 0; 26 break; 27 } 28 } 29 prim[i] = flag; 30 31 } 32 } 33 int num,f,r,target; 34 void get() 35 { 36 int n; 37 n = q[f].num; 38 39 q[f].n[4] = n%10; 40 n /= 10; 41 q[f].n[3] = n%10; 42 n /= 10; 43 q[f].n[2] = n%10; 44 n /= 10; 45 q[f].n[1] = n; 46 } 47 int getsum(struct node temp) 48 { 49 return temp.n[1]*1000+temp.n[2]*100+temp.n[3]*10+temp.n[4]; 50 } 51 void bfs() 52 { 53 int i,j; 54 memset(vis,0,sizeof(vis)); 55 f = r = 0; 56 q[r].num = num; 57 q[r].step = 0; 58 get(); 59 r++; 60 vis[num] = 1; 61 leap = 0; 62 while(f < r) 63 { 64 struct node temp,now; 65 get(); 66 now = q[f]; 67 now.step = q[f].step+1; 68 f++; 69 for(i = 0;i < 10;i++) 70 { 71 for(j = 1;j < 5;j++) 72 { 73 temp = now; 74 if(i != temp.n[j]) 75 { 76 temp.n[j] = i; 77 temp.num = getsum(temp); 78 79 if(!vis[temp.num] && prim[temp.num]) 80 { 81 vis[temp.num] = 1; 82 q[r] = temp; 83 if(temp.num == target) 84 { 85 leap = 1; 86 break; 87 } 88 r++; 89 } 90 } 91 } 92 if(leap) 93 break; 94 } 95 if(leap) 96 break; 97 } 98 return ; 99 } 100 int main() 101 { 102 int t; 103 make_prim(); 104 scanf("%d",&t); 105 106 while(t--) 107 { 108 scanf("%d %d",&num,&target); 109 if(num == target) 110 { 111 puts("0"); 112 continue; 113 } 114 115 bfs(); 116 if(leap) 117 printf("%d\n",q[r].step); 118 else 119 printf("Impossible\n"); 120 121 } 122 123 return 0; 124 }