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 ",dp[n]); 79 80 } 81 return 0; 82 }
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 ",ans.mat[0][tn]); 101 } 102 return 0;