Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5421 Accepted Submission(s):
1286
General: the generals can move and capture
one point either vertically or horizontally and cannot leave the
“palace” unless the situation called “flying general” (see the
figure above). “Flying general” means that one general can “fly” across the
board to capture the enemy general if they stand on the same line without
intervening pieces.
Chariot: the chariots can move and capture
vertically and horizontally by any distance, but may not jump over intervening
pieces
Cannon: the cannons move like the chariots,
horizontally and vertically, but capture by jumping exactly one piece
(whether it is friendly or enemy) over to its target.
Horse: the horses
have 8 kinds of jumps to move and capture shown in the left figure. However, if
there is any pieces lying on a point away from the horse horizontally or
vertically it cannot move or capture in that direction (see the figure below),
which is called “hobbling the horse’s leg”.
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 int vis[15][15]; 7 bool search(int a,int b) ///如果将军在此点会被攻击，返回0，否则返回1 8 { 9 int i,j,t; 10 if(a-2>0&&b-1>0&&vis[a-2][b-1]==5&&!vis[a-1][b-1]) return 0; ///先判断是否有能攻击这个地方的马，注意马不能撇脚 11 if(b-2>0&&a-1>0&&vis[a-1][b-2]==5&&!vis[a-1][b-1]) return 0; 12 if(a-2>0&&b+1<=9&&vis[a-2][b+1]==5&&!vis[a-1][b+1]) return 0; 13 if(b+2<=9&&a-1>0&&vis[a-1][b+2]==5&&!vis[a-1][b+1]) return 0; 14 if(a+2<=10&&b+1<=9&&vis[a+2][b+1]==5&&!vis[a+1][b+1]) return 0; 15 if(b-2>0&&a+1<=10&&vis[a+1][b-2]==5&&!vis[a+1][b-1]) return 0; 16 if(a+2<=10&&b-1>0&&vis[a+2][b-1]==5&&!vis[a+1][b-1]) return 0; 17 if(b+2<=9&&a+1<=10&&vis[a+1][b+2]==5&&!vis[a+1][b+1]) return 0; 18 t=0; 19 for(i=a-1; i>=1; i--) ///搜将军的上下左右的四个方向是否有可以攻击将军的棋子， 20 { 21 if(!t&&(vis[i][b]==3||vis[i][b]==2)) return 0; ///碰到车或者帅，可攻击此点 22 23 if(vis[i][b]==4&&t==1)return 0; ///碰到前方有炮架子的炮， 24 if(vis[i][b]!=0) t++; ///所有的棋可充当炮架子 25 } 26 t=0; 27 for(i=a+1; i<=10; i++) 28 { 29 if(!t&&(vis[i][b]==3||vis[i][b]==2)) return 0; 30 31 if(vis[i][b]==4&&t==1) return 0; 32 if(vis[i][b]!=0) t++; 33 } 34 t=0; 35 for(i=b-1; i>=1; i--) 36 { 37 if(!t&&(vis[a][i]==3||vis[a][i]==2)) return 0; 38 39 if(vis[a][i]==4&&t==1) return 0; 40 if(vis[a][i]!=0) t++; 41 } 42 t=0; 43 for(i=b+1; i<=9; i++) 44 { 45 if(!t&&(vis[a][i]==3||vis[a][i]==2)) return 0; 46 47 if(vis[a][i]==4&&t==1) return 0; 48 if(vis[a][i]!=0) t++; 49 } 50 return 1; ///将军在此点不会被攻击到，返回1 51 } 52 int main() 53 { 54 int n,xx,yy,i,j,aa,bb,x,y; 55 char c; 56 while(~scanf("%d%d%d",&n,&xx,&yy)) 57 { 58 aa=0,bb=0; 59 if(n==0&&xx==0&&yy==0) 60 break; 61 memset(vis,0,sizeof(vis)); 62 for(i=1; i<=n; i++) 63 { 64 // getchar(); 65 // scanf("%c %d %d",&c,&x,&y); 66 cin>>c>>x>>y; ///scanf输入容易WA 67 //cout<<c<<endl; 68 if(c==‘G‘) ///直接用V的值记录棋子，初始为0，帅为2，车为3，炮为4，马为5 69 { 70 vis[x][y]=2; 71 aa=x,bb=y; 72 } 73 else if(c==‘R‘) 74 vis[x][y]=3; 75 else if(c==‘C‘) 76 vis[x][y]=4; 77 else if(c==‘H‘) 78 vis[x][y]=5; 79 } 80 int k; 81 /* if(aa!=0&&bb!=0) 82 if(yy==bb) 83 { 84 for(i=xx+1; i<aa; i++) 85 if(vis[i][bb]!=0) 86 break; 87 if(i==aa) 88 { 89 printf("NO\n"); 90 continue; 91 } 92 }*/ 93 if(xx+1<=3&&xx+1>=1) ///将军不能超出边界 94 { 95 k=search(xx+1,yy); 96 if(k) ///出现1，则将军起码有一个点安全，所以未将死，输出NO 97 { 98 printf("NO\n"); 99 continue; 100 } 101 } 102 103 if(xx-1>=1&&xx-1<=3) 104 { 105 k=search(xx-1,yy); 106 if(k) 107 { 108 printf("NO\n"); 109 continue; 110 } 111 } 112 if(yy+1<=6&&yy+1>=4) 113 { 114 k=search(xx,yy+1); 115 if(k) 116 { 117 printf("NO\n"); 118 continue; 119 } 120 } 121 if(yy-1>=4&&yy-1<=6) 122 { 123 k=search(xx,yy-1); 124 if(k) 125 { 126 printf("NO\n"); 127 continue; 128 } 129 } 130 printf("YES\n"); ///没有出现为将死的情况，输出YES 131 } 132 return 0; 133 }
原文：http://www.cnblogs.com/pshw/p/5527709.html