pku3670

pku3670

http://acm.pku.edu.cn/JudgeOnline/problem?id=3670

 1  #include  < iostream >
 2  #include  < vector >
 3 
 4  using   namespace  std;
 5  const   int  N = 30005 ;
 6  int  dp1[N];
 7  int  dp2[N];
 8  int  dp3[N];
 9  int  d1[N];
10  int  d2[N];
11  int  main()
12  {
13       int  n;
14       while (cin >> n){
15           int  num;
16          memset(dp1, 0 , sizeof (dp1));
17          memset(dp2, 0 , sizeof (dp2));
18          memset(dp3, 0 , sizeof (dp3));
19           for ( int  i = 0 ;i < n;i ++ ){
20              cin >> d1[i];
21              d2[n - i - 1 ] = d1[i];
22          }
23           switch (d1[ 0 ]){
24               case   1 :
25                  dp1[ 0 ] = 0 ,dp2[ 0 ] = 1 ,dp3[ 0 ] = 1 ;
26                   break ;
27               case   2 :
28                  dp1[ 0 ] = 1 ,dp2[ 0 ] = 0 ,dp3[ 0 ] = 1 ;
29                   break ;
30               case   3 :
31                  dp1[ 0 ] = 1 ,dp2[ 0 ] = 1 ,dp3[ 0 ] = 0 ;
32                   break ;
33          }
34 
35           for ( int  i = 1 ;i < n;i ++ ){
36               switch (d1[i]){
37                   case   1 :
38                      dp1[i] = dp1[i - 1 ];
39                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]) + 1 ;
40                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ])) + 1 ;
41                       break ;
42                   case   2 :
43                      dp1[i] = dp1[i - 1 ] + 1 ;
44                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]);
45                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ])) + 1 ;
46                       break ;
47                   case   3 :
48                      dp1[i] = dp1[i - 1 ] + 1 ;
49                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]) + 1 ;
50                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ]));
51                       break ;
52              }
53          }
54           int  ans = 0x7fffffff ;
55           if (ans > dp1[n - 1 ])ans = dp1[n - 1 ];
56           if (ans > dp2[n - 1 ])ans = dp2[n - 1 ];
57           if (ans > dp3[n - 1 ])ans = dp3[n - 1 ];
58 
59           switch (d2[ 0 ]){
60               case   1 :
61                  dp1[ 0 ] = 0 ,dp2[ 0 ] = 1 ,dp3[ 0 ] = 1 ;
62                   break ;
63               case   2 :
64                  dp1[ 0 ] = 1 ,dp2[ 0 ] = 0 ,dp3[ 0 ] = 1 ;
65                   break ;
66               case   3 :
67                  dp1[ 0 ] = 1 ,dp2[ 0 ] = 1 ,dp3[ 0 ] = 0 ;
68                   break ;
69          }
70 
71           for ( int  i = 1 ;i < n;i ++ ){
72               switch (d2[i]){
73                   case   1 :
74                      dp1[i] = dp1[i - 1 ];
75                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]) + 1 ;
76                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ])) + 1 ;
77                       break ;
78                   case   2 :
79                      dp1[i] = dp1[i - 1 ] + 1 ;
80                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]);
81                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ])) + 1 ;
82                       break ;
83                   case   3 :
84                      dp1[i] = dp1[i - 1 ] + 1 ;
85                      dp2[i] = min(dp1[i - 1 ],dp2[i - 1 ]) + 1 ;
86                      dp3[i] = min(dp1[i - 1 ],min(dp2[i - 1 ],dp3[i - 1 ]));
87                       break ;
88              }
89          }
90 
91           if (ans > dp1[n - 1 ])ans = dp1[n - 1 ];
92           if (ans > dp2[n - 1 ])ans = dp2[n - 1 ];
93           if (ans > dp3[n - 1 ])ans = dp3[n - 1 ];
94 
95          cout << ans << endl;
96      }
97       return   0 ;
98  }
99