2014 ACM/ICPC Asia Regional Xi'an Online

03 hdu5009

状态转移方程很好想，dp[i] = min(dp[j]+o[j~i]^2,dp[i]) ,o[j~i]表示从j到i颜色的种数。

普通的O（n*n）是会超时的，可以想到o[]最大为sqrt(n),问题是怎么快速找到从i开始往前2种颜色、三种、四种。。。o[]种的位置。

离散化之后，可以边走边记录某个数最后一个出现的位置，初始为-1，而所要求的位置就等于

if(last[a[i]]==-1) 该数没有出现过，num[i][1] = i,num[i][j+1] = num[i-1][j];

else  last[a[i]]之前 num[i][1] = i,num[i][j+1] = num[i-1][j],之后num[i][j]= num[i-1][j];

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 #include<map>

11 using namespace std;

12 #define N 50010

13 #define LL long long

14 #define INF 0xfffffff

15 const double eps = 1e-8;

16 const double pi = acos(-1.0);

17 const double inf = ~0u>>2;

18 int a[N];

19 int dp[N];

20 int num[2][300],last[N];

21 map<int,int>f;

22 int main()

23 {

24     int i,j,n;

25     while(scanf("%d",&n)!=EOF)

26     {

27         f.clear();

28         int g =0 ;

29         for(i = 1; i<= n ;i++)

30         {

31             scanf("%d",&a[i]);

32             if(!f[a[i]])

33             {

34                 f[a[i]] = ++g;

35                 a[i] = g;

36             }

37             else a[i] = f[a[i]];

38         }

39         for(i = 1; i <= n; i++)

40         dp[i] = INF;

41         memset(last,-1,sizeof(last));

42         memset(num,0,sizeof(num));

43         int k = sqrt(n*1.0)+1;

44         int tk = 1;

45         dp[1] = 1;

46         last[a[1]] = 1;

47         num[1][1] = 1;

48         dp[0] = 0;

49         for(i = 2; i <= n ;i++)

50         {

51             if(last[a[i]]==-1)

52             {

53                 tk+=1;

54                 num[i%2][1] = i;

55                 for(j = 1; j <= min(tk-1,k-1) ; j++)

56                 num[i%2][j+1] = num[(i-1)%2][j];

57             }

58             else

59             {

60 

61                 num[i%2][1] = i;

62                 for(j = 1; j < min(k,tk) ; j++)

63                 {

64                     if(last[a[i]]==num[(i-1)%2][j]) break;

65                     num[i%2][j+1] = num[(i-1)%2][j];

66                 }

67                 for(int g = j+1 ; g <= min(tk,k) ; g++)

68                 num[i%2][g] = num[(i-1)%2][g];

69             }

70             last[a[i]] = i;

71             for(j = 1; j <= min(k,tk); j++)

72             {

73                 int po = num[i%2][j+1];

74                 dp[i] = min(dp[i],dp[po]+j*j);

75                // cout<<dp[po]<<" "<<po<<endl;

76             }

77         }

78         printf("%d\n",dp[n]);

79 

80     }

81     return 0;

82 }

View Code

 

09 hdu5015

构造矩阵

先构造出1*(n+2)的矩阵 （233, 233+a1, 233+a1+a2, 233+a1+a2+a3, ..., 233+a1+a2+..+an, 1),表示第一列上的值。

此矩阵为A，然后想要使A*B = C，C为第二列的值，所以B可以为

10 10 10 10 10 .......0

0   1    1   1   1........0

0   0    1   1   1........0

0   0    0   1   1........0

0   0    0   0   1........0

3   3    3   3   3........1

然后快速幂就可以了。。

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<vector>

  7 #include<cmath>

  8 #include<queue>

  9 #include<set>

 10 using namespace std;

 11 #define N 12

 12 #define LL __int64

 13 #define INF 0xfffffff

 14 const double eps = 1e-8;

 15 const double pi = acos(-1.0);

 16 const double inf = ~0u>>2;

 17 #define mod 10000007

 18 struct Mat

 19 {

 20     LL mat[N][N];

 21 };

 22 int n;

 23 int a[N];

 24 Mat operator * (Mat a,Mat b)

 25 {

 26     Mat c;

 27     memset(c.mat,0,sizeof(c.mat));

 28     int i,j,k;

 29     for(k =0 ; k < n ; k++)

 30     {

 31         for(i = 0 ; i < n ; i++)

 32         {

 33             if(a.mat[i][k]==0) continue;//优化

 34             for(j = 0 ; j < n ; j++)

 35             {

 36                 if(b.mat[k][j]==0) continue;//优化

 37                 c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%mod)%mod;

 38             }

 39         }

 40     }

 41     return c;

 42 }

 43 Mat operator ^(Mat a,int k)

 44 {

 45     Mat c;

 46     int i,j;

 47     for(i =0 ; i < n ; i++)

 48         for(j = 0; j < n ; j++)

 49             c.mat[i][j] = (i==j);

 50     for(; k ; k >>= 1)

 51     {

 52         if(k&1) c = c*a;

 53         a = a*a;

 54     }

 55     return c;

 56 }

 57 int main()

 58 {

 59     int m,i,j;

 60     Mat c;

 61     while(scanf("%d%d",&n,&m)!=EOF)

 62     {

 63         memset(c.mat,0,sizeof(c.mat));

 64         for(i = 0 ; i <= n; i++)

 65         {

 66             c.mat[0][i] = 10;

 67             for(j = 1 ; j <= i ; j++)

 68                 c.mat[j][i] = 1;

 69         }

 70         for(i = 0 ; i <= n ; i++)

 71             c.mat[n+1][i] = 3;

 72         c.mat[n+1][n+1] = 1;

 73         LL s = 233;

 74         Mat ans;

 75         memset(ans.mat,0,sizeof(ans));

 76         ans.mat[0][0] = s;

 77         for(i = 1; i <= n; i++)

 78         {

 79             scanf("%d",&a[i]);

 80             s+=a[i];

 81             s%=mod;

 82             ans.mat[0][i] = s;

 83         }

 84         ans.mat[0][n+1] = 1;

 85         int tn = n;

 86         n+=2;

 87         if(m>1)

 88         ans = ans*(c^(m-1));

 89 //        for(i = 0 ; i < n ; i++)

 90 //            {for(j = 0; j< n ; j++)

 91 //        cout<<c.mat[i][j]<<" ";

 92 //        puts("");

 93 //            }

 94 //        for(i = 0 ; i < n ; i++)

 95 //        {

 96 //            for(j = 0; j < n ; j++)

 97 //                cout<<ans.mat[i][j]<<" ";

 98 //            puts("");

 99 //        }

100         printf("%I64d\n",ans.mat[0][tn]);

101     }

102     return 0;

View Code