A.Xiangqi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 924 Accepted Submission(s): 244
Now we introduce some basic rules of Xiangqi. Xiangqi is played on a 10×9 board and the pieces are placed on the intersections (points). The top left point is (1,1) and the bottom right point is (10,9). There are two groups of pieces marked by black or red Chinese characters, belonging to the two players separately. During the game, each player in turn moves one piece from the point it occupies to another point. No two pieces can occupy the same point at the same time. A piece can be moved onto a point occupied by an enemy piece, in which case the enemy piece is "captured" and removed from the board. When the general is in danger of being captured by the enemy player on the enemy player’s next move, the enemy player is said to have "delivered a check". If the general's player can make no move to prevent the general's capture by next enemy move, the situation is called “checkmate”.
We only use 4 kinds of pieces introducing as follows:
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”.
Now you are given a situation only containing a black general, a red general and several red chariots, cannons and horses, and the red side has delivered a check. Now it turns to black side’s move. Your job is to determine that whether this situation is “checkmate”.
There is a blank line between two test cases. The input ends by 0 0 0.
题目大意：
判断黑方是否为死棋。
思路：
果断模拟之。。。
我将一个子的杀点(该能杀到的点)和所占点(该子所在的位置)分开保存。
马注意鳖马腿.处理方法:向一个方向跳的时候，记录蹩马腿的位置，加上判断此位置是否是某子的所占点。
注意：
车的所占点，并不是他的杀点！！
帅、车的杀点，包括挡在他前面的子的所在点。
炮的杀点，包括挡在炮前面的前面的子的所在点。
这次真的感受到了出数据的力量。上面注意的三点，自己全都没考虑到。而且细节方面考虑太不周到了，调试出来后都是写不该有的失误。
模拟题还是要多练啊！！！。。。静下心，把每一步都确认好。。。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
const int inf=0x7fffffff;
using namespace std;
struct point//red pieces
{
int x;
int y;
char va;
}p[8];
int n;
int fx,fy;//black general坐标
int vis[10][9];//记录棋盘坐标上是否有red pieces
int kill[10][9];//记录轮到red走时的杀点
int hor[8][2]={{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2}};//马走日
int hobo[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
//判断马的蹩脚点，black pieces能走的四个方向
bool noover(int x,int y)//坐标是否在棋盘内
{
if ((x>=0)&&(x<10)&&(y>=0)&&(y<9))
return true;
else return false;
}
bool gnoover(int x,int y)//坐标是否在general能落的田字格内
{
if ((x>=0)&&(x<3)&&(y>=3)&&(y<6))
return true;
else return false;
}
void ban(point p)//找出red pieces各个可以走的杀点。
{
if ((p.va=='G')||(p.va=='R'))
{
for (int i=p.x+1;i<10;i++)//十字四个方向
{
kill[i][p.y]=1;
if (vis[i][p.y]==1)
break;
}
for (int i=p.x-1;i>=0;i--)
{
kill[i][p.y]=1;
if (vis[i][p.y]==1)
break;
}
for (int i=p.y+1;i<9;i++)
{
kill[p.x][i]=1;
if (vis[p.x][i]==1)
break;
}
for (int i=p.y-1;i>=0;i--)
{
kill[p.x][i]=1;
if (vis[p.x][i]==1)
break;
}
}
else if(p.va=='H')
{
for (int i=0;i<8;i++)
if (noover(p.x+hobo[i/2][0],p.y+hobo[i/2][1]))
if (vis[p.x+hobo[i/2][0]][p.y+hobo[i/2][1]]==0)
{
if (noover(p.x+hor[i][0],p.y+hor[i][1]))
kill[p.x+hor[i][0]][p.y+hor[i][1]]=1;
}
}
else if(p.va=='C')
{
for (int i=p.x+1;i<10;i++)
//十字四个方向
if (vis[i][p.y]==1)
{
for (int j=i+1;j<10;j++)
{
kill[j][p.y]=1;
if (vis[j][p.y]==1)
break;
}
break;
}
for (int i=p.x-1;i>=0;i--)
if (vis[i][p.y]==1)
{
for (int j=i-1;j>=0;j--)
{
kill[j][p.y]=1;
if (vis[j][p.y]==1)
break;
}
break;
}
for (int i=p.y+1;i<9;i++)
if (vis[p.x][i]==1)
{
for (int j=i+1;j<9;j++)
{
kill[p.x][j]=1;
if (vis[p.x][j]==1)
break;
}
break;
}
for (int i=p.y-1;i>=0;i--)
if (vis[p.x][i]==1)
{
for (int j=i-1;j>=0;j--)
{
kill[p.x][j]=1;
if (vis[p.x][j]==1)
break;
}
break;
}
}
}
int main()
{
while(cin>>n>>fx>>fy)
{
if ((n&&fx&&fy)==0) return 0;
fx--;
fy--;
memset(vis,0,sizeof(vis));
memset(kill,0,sizeof(kill));
for (int i=0;i<n;i++)
{
cin>>p[i].va>>p[i].x>>p[i].y;
p[i].x-=1;
p[i].y-=1;
vis[p[i].x][p[i].y]=1;
}
for (int i=0;i<n;i++)
ban(p[i]);
//for (int i=0;i<10;i++)
// for (int j=0;j<9;j++)
// cout<<i<<' '<<j<<' '<<kill[i][j]<<endl;
int ok=0;
for (int i=0;i<4;i++)
if (gnoover(fx+hobo[i][0],fy+hobo[i][1]))
if (!kill[fx+hobo[i][0]][fy+hobo[i][1]])
ok=1;
if (ok)
cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return 0;
}