ZOJ 2610 Puzzle (YY)

题意：一个又n^2个小三角形组成的三角形又A，B两块三角形组成，问能否把它们分开
解题思路：平行于三条边移动。
解题代码：
  1 // File Name: g.c
  2 // Author: darkdream
  3 // Created Time: 2013年09月09日 星期一 13时45分49秒
  4 
  5 #include<stdio.h>
  6 #include<string.h>
  7 #include<stdlib.h>
  8 #include<time.h>
  9 #include<math.h>
 10 #define LL long long
 11 
 12 //freopen("/home/plac/problem/input.txt","r",stdin);
 13 //freopen("/home/plac/problem/output.txt","w",stdout);
 14 int a[200][200];
 15 int n ;
 16 int ok1()
 17 {
 18        for(int i = 1;i <= n;i ++)
 19        {
 20          int k = 2*i-1;
 21          for(int j = 1;j <= k; j ++)
 22          if(a[i][j] == 0 && a[i][j-1] == 1)
 23          {
 24               k = 0 ;
 25               return 0;
 26          }
 27        }
 28        return 1; 
 29 }
 30 int ok2()
 31 {
 32        for(int i = 1;i <= n;i ++)
 33        {
 34          int k = 2*i-1;
 35          for(int j = 1; j <= k; j++)
 36          if(a[i][j] == 0 && a[i][j+1] == 1)
 37          {
 38               return 0;
 39          }
 40        }
 41        return 1; 
 42 }
 43 int ok3()
 44 {
 45        for(int i = 1;i <= n;i ++)
 46        {
 47          int k = 2*i-1;
 48          for(int j = 1; j <= k; j++)
 49          if(a[i][j] == 0 )
 50          {
 51               if(j % 2 == 1)
 52               {
 53                  if(a[i][j-1] == 1)
 54                     return 0 ;
 55               }
 56               else
 57               {
 58                   if(a[i-1][j-1] == 1)
 59                       return 0;
 60               }
 61          }
 62        }
 63        return 1; 
 64 }
 65 int ok4()
 66 {
 67        for(int i = 1;i <= n;i ++)
 68        {
 69          int k = 2*i-1;
 70          for(int j = 1;  j <= k; j ++)
 71          if(a[i][j] == 0 )
 72          {
 73               if(j % 2 == 1)
 74               {
 75                  if(a[i+1][j+1] == 1)
 76                     return 0 ;
 77               }
 78               else
 79               {
 80                   if(a[i][j+1] == 1)
 81                       return 0;
 82               }
 83          }
 84        }
 85        return 1; 
 86 }
 87 int ok5()
 88 {
 89        for(int i = 1;i <= n;i ++)
 90        {
 91          int k = 2*i-1;
 92          for(int j = 1; j <= k ; j ++)
 93          if(a[i][j] == 0 )
 94          {
 95               if(j % 2 == 1)
 96               {
 97                  if(a[i][j+1] == 1)
 98                     return 0 ;
 99               }
100               else
101               {
102                   if(a[i-1][j-1] == 1)
103                       return 0;
104               }
105          }
106        }
107        return 1; 
108 }
109 int ok6()
110 {
111        for(int i = 1;i <= n;i ++)
112        {
113          int k = 2*i-1;
114          for(int j = 1; j <= k; j ++)
115          if(a[i][j] == 0 )
116          {
117               if(j % 2 == 1)
118               {
119                  if(a[i+1][j+1] == 1)
120                     return 0 ;
121               }
122               else
123               {
124                   if(a[i][j-1] == 1)
125                       return 0;
126               }
127          }
128        }
129        return 1; 
130 }
131 int main(){
132      int t  = 0 ;
133      while(scanf("%d",&n)!= EOF && n)
134      {
135        int ok = 0 ;
136        memset(a,0,sizeof(a));
137        t++;
138        for(int i = 1;i <= n;i ++)
139        {
140          int k = 2*i-1;
141          for(int j= 1 ;j <= k; j ++ )
142              scanf("%1d",&a[i][j]);
143        }
144          if(ok1()||ok2()||ok3()||ok4()||ok5()||ok6())
145              ok = 1;
146          printf("Puzzle %d\n",t);
147          if(ok)
148              printf("Parts can be separated\n");
149          else printf("Parts cannot be separated\n");
150 
151      }
152 
153 return 0 ;
154 }
View Code
posted on 2013-09-09 23:43 dark_dream 阅读(362) 评论(0) 编辑收藏举报