Xiangqi UVa1589（多次WA）

  1 #include<stdio.h>
  2 #include<string.h>
  3 #define maxn 8
  4 //#define LOCAL
  5 #include<iostream>
  6 using namespace std;
  7 int test();
  8 int count(int,int);
  9 int horse_judge(int);
 10 int hobble(int,int);
 11 
 12 int G1, G2;   //(G1,G2) of black general.
 13 int N;        //the number of pieces of red.
 14 char a[maxn];  //store the pieces name in red.         # i  to (2*i,2*i+1)
 15 int pos[2*maxn]; //record the point.
 16 
 17 //judge YES(1) or NO(0)
 18 int judgement()
 19 {   //if the black and red general in the same line and do not have pieces between them.
 20     //black win.
 21     int index = strchr(a,‘G‘)-a;
 22     if( pos[2*index+1] == G2 && count(2,index) == 0) {return 0;}
 23 
 24     //move up, down, left, right to try if the black general can save itself.
 25     //if this move is a life way, the program will return 0;
 26     //else the program will go back to the original position and then try different directions.
 27     if( G1-1 >= 1 ){ G1 -= 1; if(test()) return 0; G1 += 1; }
 28 
 29     if( G1+1 <= 3 ){ G1 += 1; if(test()) return 0; G1 -= 1; }
 30 
 31     if( G2-1 >= 4 ){ G2 -= 1; if(test()) return 0; G2 += 1; }
 32 
 33     if( G2+1 <= 6 ){ G2 += 1; if(test()) return 0; G2 -= 1; }
 34     return 1;
 35 }
 36 
 37 //test if the black general will die(4 possibilities).
 38 //when the black general do not die, the program will return 1
 39 //when die, return 0
 40 int test()
 41 {
 42     int n = strlen(a);
 43     for(int i = 0; i < n; i++)
 44     {
 45         //if two generals in the same line, I will count the number of pieces between them.
 46         //if the number is zero, I‘m sorry to say that the black general die.
 47         if( a[i] == ‘G‘ && G2 == pos[2*i+1] )
 48         {
 49             if( count(2,i) == 0 )  return 0;
 50         }
 51 
 52         if( a[i] == ‘R‘ && (G1 == pos[2*i] || G2 == pos[2*i+1]) )
 53         {
 54             int flag = 0; //used to judge if they are in the same row or col.
 55             if(G1 == pos[2*i] && G2 == pos[2*i+1])continue;//eat the piece.
 56             if(G1 == pos[2*i]) flag = 1;
 57             if(G2 == pos[2*i+1]) flag = 2;
 58             if( count(flag,i) == 0 ) { return 0;}
 59         }
 60         if( a[i] == ‘C‘ && (G1 == pos[2*i] || G2 == pos[2*i+1]) )
 61         {
 62             int flag = 0;
 63             if(G1 == pos[2*i] && G2 == pos[2*i+1]) continue;
 64             if(G1 == pos[2*i]) flag = 1;
 65             if(G2 == pos[2*i+1]) flag = 2;
 66             if( count(flag,i) == 1 ) { return 0;}
 67 
 68         }
 69         if( a[i] == ‘H‘ )
 70         {
 71             if(G1 == pos[2*i] && G2 == pos[2*i+1]) continue;
 72             if(horse_judge(i)){ return 0; } //judge by other functions.
 73         }
 74     }
 75     return 1;//if the black survive after all the test, return 1;
 76 }
 77 
 78 //count the number of other pieces between these two pieces.
 79 //1 is row; 2 is col
 80 int count(int flag, int index)
 81 {
 82     int n = strlen(a);
 83     int num = 0;
 84     int t(0);
 85     //count the row.
 86     if(flag == 1)
 87     {
 88         for(int i = 0; i < n; i++ )
 89         {
 90             if(pos[2*i] == G1 ) 
 91             {
 92                 //on the left.
 93                 if(pos[2*i+1]<G2 && (pos[2*i+1] > pos[2*index+1]))
 94                     num++;
 95                 //on the right.
 96                 if(pos[2*i+1]>G2 && (pos[2*i+1] < pos[2*index+1]))
 97                     num++;
 98             }
 99         }
100         return num;
101     }
102     //count the column.
103     if(flag == 2)
104     {
105         for(int i = 0; i < n; i++)
106         {
107             if(pos[2*i+1] == G2)
108             {
109                 //above the  of general
110                 if(pos[2*i] < G1 && pos[2*i] > pos[2*index])
111                     num++;
112                 //under the general
113                 if(pos[2*i] > G1 && pos[2*i] < pos[2*index])
114                     num++;
115             }
116         }
117         return num;
118     }
119 }
120 
121 //judge if the black general will be killed by horse.
122 //return 0 live.
123 //return 1 die.
124 int horse_judge(int index)
125 {
126     //cx,cy are uesed to know the horse‘s direction to the black general.
127     int cx, cy;
128     int x(pos[index*2]), y(pos[index*2+1]); //the point of the horse.
129     if( G1 > x ) cx = -1;
130     else if( G1 < x ) cx = 1;
131     else return 0;
132     if( G2 > y ) cy = -1;
133     else if( G2 < y ) cy = 1;
134     else return 0;
135     if( (G1+ cx*1 == x && G2 + cy*2 == y)|| (G1 + cx*2 == x && G2 + cy*1 == y) )
136     {
137         int point1 = G1+1*cx;
138         int point2 = G2+1*cy;
139         if(hobble(point1, point2) ) return 0;
140         else return 1;
141     }
142     else return 0;    
143 }
144 
145 int hobble(int x, int y)
146 {
147     for(int i = 0; i < strlen(a); i++)
148     {
149         if( pos[2*i]==x && pos[2*i+1] == y) return 1;
150     }
151     return 0;
152 }
153 
154 int main()
155 {
156     #ifdef LOCAL
157     freopen("data.in", "r", stdin);
158     freopen("data.out", "w", stdout);
159     #endif
160 
161     while( cin>>N>>G1>>G2 && N != 0)     //!!!!!!!!!!!!!!!!
162     {
163         //initialize two arraies.
164         memset(a,0,sizeof(a));
165         memset(pos,0,sizeof(pos));
166         for(int i = 0; i < N; i++)
167         {
168             cin>>a[i];
169             cin>>pos[2*i]>>pos[2*i+1];
170         }
171         if(judgement()) 
172             printf("YES\n");    
173         else 
174             printf("NO\n");
175     }
176     return 0;
177 }