POJ Xiangqi 4001 && HDOJ 4121 Xiangqi

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题目链接(POJ):http://poj.org/problem?id=4001

题目链接(HDOJ):http://acm.hdu.edu.cn/showproblem.php?pid=4121

Description

Xiangqi is one of the most popular two-player board games in China. The game represents a battle between two armies with the goal of capturing the enemy’s “general” piece. In this problem, you are given a situation of later stage in the game. Besides, the red side has already “delivered a check”. Your work is to check whether the situation is “checkmate”. 



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 right figure). “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 left figure), which is called “hobbling the horse’s leg”.
  

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”.

Input

Output

Sample Input

2 1 4
G 10 5
R 6 4

3 1 5
H 4 5
G 10 5
C 7 5

0 0 0

Sample Output

YES

NO
Hint
In the first situation, the black general is checked by chariot and “flying general”. In the second situation, the black general can move to (1, 4) or (1, 6) to stop check. See the figure above. 


Situation 1


Situation 2
Source
Fuzhou 2011

题意:给出红子和黑子的位置，判断黑子是否被将死。

题解:模拟题，根据各个棋子的行走规则，标记"杀点"即可。最后判断黑子将的活动范围是否全都在"杀点"范围内。

PS: 还是好几个月前做的这道题，一直WA。今天来处理下这个遗留问题，还很顺利，一次AC。这种题主要是思路清晰and细心

AC代码:

#include<iostream>
#include<cstring>
#define N 11
using namespace std;
int n,gx,gy,x,y,maze[N][N],pie[N][2];
char ch,chess[N][N];
int hdir[8][2]={
    {-2,-1},{-2,1},{-1,2},{1,2},
    {2,-1},{2,1},{-1,-2},{1,-2}
};
int Isdir[4][2]={
    {-1,0},{0,1},{1,0},{0,-1}
};
bool In(int dx,int dy){
    if(dx>0&&dx<11&&dy>0&&dy<11)return true;
    return false;
}
void SetFlag(int dx,int dy){
    int tx,ty;
    switch(chess[dx][dy]){
    case 'G':
        tx=dx-1;
        while(tx>0&&chess[tx][dy]==0){
            maze[tx][dy]=0;tx--;
        }
        if(tx>0)maze[tx][dy]=0;
        break;
    case 'H':
        for(int i=0;i<8;i++){
            tx=dx+hdir[i][0],ty=dy+hdir[i][1];
            if(In(tx,ty)&&!chess[Isdir[i/2][0]+dx][Isdir[i/2][1]+dy]){
                maze[tx][ty]=0;
            }
        }
        break;
    case 'C':
        for(int i=0;i<4;i++){
            tx=Isdir[i][0]+dx,ty=Isdir[i][1]+dy;
            while(In(tx,ty)&&!chess[tx][ty]){
                tx+=Isdir[i][0];ty+=Isdir[i][1];
            }
            tx+=Isdir[i][0];ty+=Isdir[i][1];
            while(In(tx,ty)&&!chess[tx][ty]){
                maze[tx][ty]=0;
                tx+=Isdir[i][0];ty+=Isdir[i][1];
            }
        }
        break;
    case 'R':
        for(int i=0;i<4;i++){
            int tx=Isdir[i][0]+dx,ty=Isdir[i][1]+dy;
            while(In(tx,ty)&&!chess[tx][ty]){
                maze[tx][ty]=0;
                tx+=Isdir[i][0];ty+=Isdir[i][1];
            }
            if(In(tx,ty))maze[tx][ty]=0;
        }
        break;
    }
}
int main()
{
    cin.sync_with_stdio(false);
    while(cin>>n>>gx>>gy&&(n||gx||gy)){
        int pos=0;
        memset(maze,-1,sizeof(maze));
        memset(chess,0,sizeof(chess));
        while(n--){
            cin>>ch>>x>>y;
            pie[pos][0]=x;pie[pos++][1]=y;
            chess[x][y]=ch;
        }
        for(int i=0;i<pos;i++)
        SetFlag(pie[i][0],pie[i][1]);
        bool live=false;
        if(maze[gx][gy])live=true;
        for(int i=0;i<4;i++){
            int tx=gx+Isdir[i][0],ty=gy+Isdir[i][1];
            if(tx>0&&tx<4&&ty>3&&ty<7&&maze[tx][ty])live=true;
        }
        if(live)cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
    return 0;
}

POJ Xiangqi 4001 && HDOJ 4121 Xiangqi

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原文地址：http://blog.csdn.net/mummyding/article/details/43576895