Ali and Baba
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1284 Accepted Submission(s):
274
Problem Description
There is a rectangle area (with N rows and M columns)
in front of Ali and Baba, each grid might be one of the following:
1. Empty area, represented by an integer 0.
2. A Stone, represented by an integer x (x > 0) which denote the HP of this stone.
3. Treasure, represented by an integer -1.
Now, Ali and Baba get the map of this mysterious area, and play the following game:
Ali and Baba play alternately, with Ali starting. In each turn, the player will choose a stone that he can touch and hit it. After this operation, the HP of the stone that been hit will decrease by 1. If some stone’s HP is decreased to 0, it will become an empty area. Here, a player can touch a stone means
there is path consist of empty area from the outside to the stone. Note that two grids are adjacent if and only if they share an edge.
The player who hits the treasure first wins the game.
1. Empty area, represented by an integer 0.
2. A Stone, represented by an integer x (x > 0) which denote the HP of this stone.
3. Treasure, represented by an integer -1.
Now, Ali and Baba get the map of this mysterious area, and play the following game:
Ali and Baba play alternately, with Ali starting. In each turn, the player will choose a stone that he can touch and hit it. After this operation, the HP of the stone that been hit will decrease by 1. If some stone’s HP is decreased to 0, it will become an empty area. Here, a player can touch a stone means
there is path consist of empty area from the outside to the stone. Note that two grids are adjacent if and only if they share an edge.
The player who hits the treasure first wins the game.
Input
The input consists several testcases.
The first line contains two integer N and M (0 < N,M <= 300), the size of the maze.
The following N lines each contains M integers (less than 100), describes the maze, where a positive integer represents the HP of a stone, 0 reperents an empty area, and -1 reperents the treasure.
There is only one grid contains the treasure in the maze.
The first line contains two integer N and M (0 < N,M <= 300), the size of the maze.
The following N lines each contains M integers (less than 100), describes the maze, where a positive integer represents the HP of a stone, 0 reperents an empty area, and -1 reperents the treasure.
There is only one grid contains the treasure in the maze.
Output
“Ali Win” or “Baba Win” indicates the winner of the
game.
Sample Input
3 3
1 1 1
1 -1 1
1 1 1
Sample Output
Baba Win
Source
7 7
0 2 1 1 1 1 1
1 0 0 0 0 0 1
1 0 2 2 1 1 1
1 0 0 0 0 0 1
1 0 -1 5 3 4 1
1 0 0 0 0 0 1
1 1 1 1 1 1 1
此时ans = 9 Ali Win.
写得很挫,其实思路对的话,写起来很简单的
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<queue> 5 #include<cstdlib> 6 using namespace std; 7 8 int n,m; 9 int a[302][302]; 10 bool use[302][302]; 11 bool hash[302][302]; 12 13 int map1[4][2]={ {1,0},{0,1},{-1,0},{0,-1}}; 14 struct node 15 { 16 int x,y; 17 }; 18 queue<node>Q; 19 bool pd() 20 { 21 int i; 22 for(i=1;i<=m;i++) 23 if(use[1][i]==true) return true; 24 for(i=1;i<=m;i++) 25 if(use[n][i]==true) return true; 26 27 for(i=1;i<=n;i++) 28 if(use[i][1]==true) return true; 29 for(i=1;i<=n;i++) 30 if(use[i][m]==true) return true; 31 return false; 32 } 33 void bfs(int x,int y) 34 { 35 int i; 36 node t,cur; 37 t.x=x; 38 t.y=y; 39 Q.push(t); 40 use[x][y]=true; 41 42 while(!Q.empty()) 43 { 44 cur=Q.front(); 45 Q.pop(); 46 47 for(i=0;i<4;i++) 48 { 49 t=cur; 50 t.x=t.x+map1[i][0]; 51 t.y=t.y+map1[i][1]; 52 if(t.x>=1&&t.x<=n && t.y>=1&&t.y<=m) 53 { 54 if(use[t.x][t.y]==false && a[t.x][t.y]==0) 55 { 56 use[t.x][t.y]=true; 57 Q.push(t); 58 } 59 } 60 } 61 } 62 } 63 void dbfs(int x,int y)//second serch 64 { 65 int i; 66 node t,cur; 67 bool flag; 68 t.x=x; 69 t.y=y; 70 hash[x][y]=true; 71 Q.push(t); 72 73 while(!Q.empty()) 74 { 75 cur=Q.front(); 76 Q.pop(); 77 flag=false; 78 for(i=0;i<4;i++) 79 { 80 t=cur; 81 t.x=t.x+map1[i][0]; 82 t.y=t.y+map1[i][1]; 83 if(t.x>=1&&t.x<=n && t.y>=1&&t.y<=m) 84 { 85 if(use[t.x][t.y]==true) 86 { 87 flag=true; 88 break; 89 } 90 } 91 } 92 if(flag==true) continue; 93 for(i=0;i<4;i++) 94 { 95 t=cur; 96 t.x=t.x+map1[i][0]; 97 t.y=t.y+map1[i][1]; 98 if(t.x>=1&&t.x<=n && t.y>=1&&t.y<=m) 99 { 100 if(hash[t.x][t.y]==false && use[t.x][t.y]==false) 101 { 102 hash[t.x][t.y]=true; 103 Q.push(t); 104 } 105 } 106 } 107 } 108 } 109 void s_serch() 110 { 111 int i; 112 while(!Q.empty()) 113 { 114 Q.pop(); 115 } 116 for(i=1;i<=m;i++) 117 if(hash[1][i]==false && use[1][i]==false) 118 dbfs(1,i); 119 for(i=1;i<=m;i++) 120 if(hash[n][i]==false && use[n][i]==false) 121 dbfs(n,i); 122 123 for(i=1;i<=n;i++) 124 if(hash[i][1]==false && use[i][1]==false) 125 dbfs(i,1); 126 for(i=1;i<=n;i++) 127 if(hash[i][m]==false && use[i][m]==false) 128 dbfs(i,m); 129 } 130 int main() 131 { 132 int i,j,x,y,cur,k; 133 bool flag; 134 while(scanf("%d%d",&n,&m)>0) 135 { 136 memset(hash,false,sizeof(hash)); 137 memset(use,false,sizeof(use)); 138 for(i=1;i<=n;i++) 139 for(j=1;j<=m;j++) 140 { 141 scanf("%d",&a[i][j]); 142 } 143 while(!Q.empty()) 144 { 145 Q.pop(); 146 } 147 for(i=1;i<=n;i++) 148 for(j=1;j<=m;j++) 149 { 150 if(a[i][j]==-1 && use[i][j]==false) 151 bfs(i,j);//first serch. 152 } 153 if(pd()==true){ 154 printf("Ali Win "); 155 continue; 156 } 157 s_serch(); 158 for(i=1,cur=0;i<=n;i++) 159 for(j=1;j<=m;j++) 160 { 161 if(hash[i][j]==true && a[i][j]>0) 162 { 163 flag=false; 164 for(k=0;k<4;k++) 165 { 166 x=i+map1[k][0]; 167 y=j+map1[k][1]; 168 if(x>=1&&x<=n &&y>=1&&y<=m && use[x][y]==true) 169 { 170 flag=true; 171 break; 172 } 173 } 174 if(flag==true) cur=cur+a[i][j]-1; 175 else cur=cur+a[i][j]; 176 } 177 } 178 // printf("%d ",cur); 179 cur=cur%2; 180 if(cur==1) printf("Ali Win "); 181 else printf("Baba Win "); 182 } 183 return 0; 184 }