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Pay it forword - 我并不觉的自豪,我所尝试的事情都失败了······习惯原本生活的人不容易改变,就算现状很糟,他们也很难改变,在过程中,他们还是放弃了······他们一放弃,大家就都是输家······让爱传出去,很困难,也无法预料,人们需要更细心的观察别人,要随时注意才能保护别人,因为他们未必知道自己要什么·····
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 
posted on 2008-07-25 15:27 小果子 阅读(211) 评论(0)  编辑 收藏 引用

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