Mine
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 855 Accepted Submission(s): 259
Problem Description
Have you ever played a game in Windows: Mine? This game is played on a n*m board, just like the Pic(1) 
On the board, Under some grids there are mines (represent by a red flag). There are numbers ‘A(i,j)’ on some grids means there’re A(i,j) mines on the 8 grids which shares a corner or a line with gird(i,j). Some grids are blank means there’re no mines on the 8 grids which shares a corner or a line with them. At the beginning, all grids are back upward. In each turn, Player should choose a back upward grid to click. If he clicks a mine, Game over. If he clicks a grid with a number on it , the grid turns over. If he clicks a blank grid, the grid turns over, then check grids in its 8 directions.If the new grid is a blank gird or a grid with a number,it will be clicked too. So If we click the grid with a red point in Pic(1), grids in the area be encompassed with green line will turn over. Now Xiemao and Fanglaoshi invent a new mode of playing Mine. They have found out coordinates of all grids with mine in a game. They also find that in a game there is no grid will turn over twice when click 2 different connected components.(In the Pic(2), grid at (1,1) will turn over twice when player clicks (0,0) and (2,2) , test data will not contain these cases). Then, starting from Xiemao, they click the grid in turns. They both use the best strategy. Both of them will not click any grids with mine, and the one who have no grid to click is the loser. Now give you the size of board N, M, number of mines K, and positions of every mine Xi,Yi. Please output who will win.
Input
Multicase The first line of the date is an integer T, which is the number of the text cases. (T<=50) Then T cases follow, each case starts with 3 integers N, M, K indicates the size of the board and the number of mines.Then goes K lines, the ith line with 2 integer Xi,Yi means the position of the ith mine. 1<=N,M<=1000 0<=K<=N*M 0<=Xi<N 0<=Yi<M
Output
For each case, first you should print "Case #x: ", where x indicates the case number between 1 and T . Then output the winner of the game, either ”Xiemao” or “Fanglaoshi”. (without quotes)
Sample Input
2
3 3 0
3 3 1
1 1
Sample Output
Case #1: Xiemao
Case
#2: Fanglaoshi
Source
多校第八场的一道博弈题，说是在一个扫雷的格子中，两个人轮流点击（它们事先都知道雷的位置），转到哪一个人的时候它只能点雷，那么这个人就输了、
当时想的作法很诡异，第八场比赛本来已经抱着被AK的心态了，结果我胡乱推导的过程中发现了一丝规律，于是抱着签到（一场比赛总不能一次提交都没有吧，囧）肯必WA的心态交了一发，结果居然过了。。。
当然RE了一次，把DFS的递归调用改成了手写的栈才过了这个题，代码很丑思路也很糟，就不解释了，过几天参考解题报告想出更好的解法了再来更新
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 typedef struct node 8 { 9 int x; 10 int y; 11 } NODE; 12 13 int n,m,k; 14 int g[1050][1050],gt[1050][1050]; 15 NODE Stack[1000000]; 16 17 int main() 18 { 19 int t; 20 21 scanf("%d",&t); 22 23 for(int kase=1;kase<=t;kase++) 24 { 25 scanf("%d %d %d",&n,&m,&k); 26 memset(g,0,sizeof(g)); 27 memset(gt,0,sizeof(gt)); 28 for(int i=0;i<=n+1;i++) 29 g[i][0]=g[i][m+1]=-2; 30 for(int j=0;j<=m+1;j++) 31 g[0][j]=g[n+1][j]=-2; 32 for(int i=1;i<=k;i++) 33 { 34 int x,y; 35 scanf("%d %d",&x,&y); 36 g[x+1][y+1]=-1; 37 } 38 if(k==0) 39 { 40 printf("Case #%d: Xiemao ",kase); 41 continue; 42 } 43 for(int i=1;i<=n;i++) 44 for(int j=1;j<=m;j++) 45 { 46 if(g[i][j]==-1) 47 { 48 if(g[i-1][j-1]>=0) 49 g[i-1][j-1]++; 50 if(g[i-1][j]>=0) 51 g[i-1][j]++; 52 if(g[i-1][j+1]>=0) 53 g[i-1][j+1]++; 54 if(g[i][j-1]>=0) 55 g[i][j-1]++; 56 if(g[i][j+1]>=0) 57 g[i][j+1]++; 58 if(g[i+1][j-1]>=0) 59 g[i+1][j-1]++; 60 if(g[i+1][j]>=0) 61 g[i+1][j]++; 62 if(g[i+1][j+1]>=0) 63 g[i+1][j+1]++; 64 } 65 } 66 int boundary=0,mast=0; 67 for(int i=1;i<=n;i++) 68 for(int j=1;j<=m;j++) 69 { 70 if(g[i][j]!=0&&g[i][j]!=-1) 71 if(g[i-1][j-1]==0||g[i-1][j]==0||g[i-1][j+1]==0||g[i][j-1]==0||g[i][j+1]==0||g[i+1][j-1]==0||g[i+1][j]==0||g[i+1][j+1]==0) 72 boundary++; 73 else 74 mast++; 75 } 76 for(int i=0;i<=n+1;i++) 77 for(int j=0;j<=m+1;j++) 78 gt[i][j]=(g[i][j]==0?0:1); 79 int blocks=0; 80 for(int i=1;i<=n;i++) 81 for(int j=1;j<=m;j++) 82 { 83 if(gt[i][j]==0) 84 { 85 blocks++; 86 g[i][j]=1; 87 int l=0,r=0; 88 NODE temp; 89 temp.x=i; 90 temp.y=j; 91 Stack[r++]=temp; 92 while(l<r) 93 { 94 if(gt[Stack[l].x-1][Stack[l].y]==0) 95 { 96 gt[Stack[l].x-1][Stack[l].y]=1; 97 temp.x=Stack[l].x-1; 98 temp.y=Stack[l].y; 99 Stack[r++]=temp; 100 } 101 if(gt[Stack[l].x+1][Stack[l].y]==0) 102 { 103 gt[Stack[l].x+1][Stack[l].y]=1; 104 temp.x=Stack[l].x+1; 105 temp.y=Stack[l].y; 106 Stack[r++]=temp; 107 } 108 if(gt[Stack[l].x][Stack[l].y-1]==0) 109 { 110 gt[Stack[l].x][Stack[l].y-1]=1; 111 temp.x=Stack[l].x; 112 temp.y=Stack[l].y-1; 113 Stack[r++]=temp; 114 } 115 if(gt[Stack[l].x][Stack[l].y+1]==0) 116 { 117 gt[Stack[l].x][Stack[l].y+1]=1; 118 temp.x=Stack[l].x; 119 temp.y=Stack[l].y+1; 120 Stack[r++]=temp; 121 } 122 l++; 123 } 124 } 125 } 126 if(mast%2==1) 127 { 128 if(blocks%2==1) 129 { 130 if(boundary%2==0) 131 printf("Case #%d: Fanglaoshi ",kase); 132 else 133 printf("Case #%d: Xiemao ",kase); 134 } 135 else 136 printf("Case #%d: Xiemao ",kase); 137 } 138 else 139 { 140 if(blocks%2==1) 141 printf("Case #%d: Xiemao ",kase); 142 else 143 { 144 if(boundary%2==0) 145 printf("Case #%d: Fanglaoshi ",kase); 146 else 147 printf("Case #%d: Xiemao ",kase); 148 } 149 } 150 } 151 152 return 0; 153 }