hdu4965 Fast Matrix Calculation 矩阵快速幂

本文介绍了一种利用矩阵快速幂解决特定数学问题的方法。通过构建矩阵并进行快速幂运算,可以高效地求解大规模问题。文章详细展示了算法的实现过程,并提供了一个具体的编程示例。

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One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her.

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation.

Step 1: Calculate a new N*N matrix C = A*B.
Step 2: Calculate M = C^(N*N). 
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’.
Step 4: Calculate the sum of all the elements in M’. 

Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math.

 

矩阵快速幂

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<math.h>
  4 typedef long long ll;
  5 const int mod=6;
  6 
  7 struct mat{
  8     int r,c;
  9     int m[7][7];        //经测试最大开成590*590的 ll 型矩阵
 10     mat(){}
 11     mat(int r,int c):r(r),c(c){}
 12     void clear(){
 13         memset(m,0,sizeof(m));
 14     }
 15 
 16     mat operator+(mat a)const{
 17         mat ans(r,c);
 18         for(int i=1;i<=r;i++){
 19             for(int j=1;j<=c;j++){
 20                 ans.m[i][j]=(m[i][j]+a.m[i][j])%mod;
 21             }
 22         }
 23         return ans;
 24     }
 25 
 26     mat operator*(mat a)const{
 27         mat tmp(r,a.c);
 28         int i,j,k;
 29         for(i=1;i<=tmp.r;i++){
 30             for(j=1;j<=tmp.c;j++){
 31                 tmp.m[i][j]=0;
 32                 for(k=1;k<=c;k++){
 33                     tmp.m[i][j]=(tmp.m[i][j]+(m[i][k]*a.m[k][j])%mod)%mod;
 34                 }
 35             }
 36         }
 37         return tmp;
 38     }
 39 
 40     mat operator^(int n)const{        //需要时可以用 ll n,注意运算符优先级比较低,多用括号;
 41         mat ans(r,r),tmp(r,r);
 42         memcpy(tmp.m,m,sizeof(tmp.m));
 43         ans.clear();
 44         for(int i=1;i<=ans.r;i++){
 45             ans.m[i][i]=1;
 46         }
 47         while(n){
 48             if(n&1)ans=ans*tmp;
 49             n>>=1;
 50             tmp=tmp*tmp;
 51         }
 52         return ans;
 53     }
 54     
 55     void print()const{
 56         for(int i=1;i<=r;i++){
 57             for(int j=1;j<=c;j++){
 58                 printf("%d",m[i][j]);
 59                 if(j==c)printf("\n");
 60                 else printf(" ");
 61             }
 62         }
 63     }
 64 
 65 };
 66 
 67 int m1[2000][2000],m2[2000][2000],tmp[7][7],tmp2[2000][2000],tmp3[2000][2000];
 68 
 69 int main(){
 70     int n,k;
 71     while(scanf("%d%d",&n,&k)!=EOF&&n+k){
 72         int i,j,p;
 73         for(i=1;i<=n;i++){
 74             for(j=1;j<=k;j++)scanf("%d",&m1[i][j]);
 75         }
 76         for(i=1;i<=k;i++){
 77             for(j=1;j<=n;j++)scanf("%d",&m2[i][j]);
 78         }
 79         for(i=1;i<=k;i++){
 80             for(j=1;j<=k;j++){
 81                 tmp[i][j]=0;
 82                 for(p=1;p<=n;p++){
 83                     tmp[i][j]+=m2[i][p]*m1[p][j];
 84                 }
 85                 tmp[i][j]%=6;
 86             }
 87         }
 88         mat a(k,k);
 89         memcpy(a.m,tmp,sizeof(tmp));
 90         a=(a^(n*n-1));
 91         memcpy(tmp,a.m,sizeof(tmp));
 92         for(i=1;i<=n;i++){
 93             for(j=1;j<=k;j++){
 94                 tmp2[i][j]=0;
 95                 for(p=1;p<=k;p++){
 96                     tmp2[i][j]+=m1[i][p]*tmp[p][j];
 97                 }
 98                 tmp2[i][j]%=6;
 99             }
100         }
101         for(i=1;i<=n;i++){
102             for(j=1;j<=n;j++){
103                 tmp3[i][j]=0;
104                 for(p=1;p<=k;p++){
105                     tmp3[i][j]+=tmp2[i][p]*m2[p][j];
106                 }
107                 tmp3[i][j]%=6;
108             }
109         }
110         int ans=0;
111         for(i=1;i<=n;i++){
112             for(j=1;j<=n;j++)ans+=tmp3[i][j];
113         }
114         printf("%d\n",ans);
115     }
116     return 0;
117 }
View Code

 

转载于:https://www.cnblogs.com/cenariusxz/p/6598597.html

### HDU 2544 题目分析 HDU 2544 是关于最短路径的经典问题,可以通过多种方法解决,其中包括基于邻接矩阵的 Floyd-Warshall 算法。以下是针对该问题的具体解答。 --- #### 基于邻接矩阵的 Floyd-Warshall 实现 Floyd-Warshall 算法是一种动态规划算法,适用于计算任意两点之间的最短路径。它的时间复杂度为 \( O(V^3) \),其中 \( V \) 表示节点的数量。对于本题中的数据规模 (\( N \leq 100 \)),此算法完全适用。 下面是具体的实现方式: ```cpp #include <iostream> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; int dist[105][105]; int n, m; void floyd() { for (int k = 1; k <= n; ++k) { // 中间节点 for (int i = 1; i <= n; ++i) { // 起始节点 for (int j = 1; j <= n; ++j) { // 结束节点 if (dist[i][k] != INF && dist[k][j] != INF) { dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]); } } } } } int main() { while (cin >> n >> m && (n || m)) { // 初始化邻接矩阵 for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (i == j) dist[i][j] = 0; else dist[i][j] = INF; } } // 输入边的信息并更新邻接矩阵 for (int i = 0; i < m; ++i) { int u, v, w; cin >> u >> v >> w; dist[u][v] = min(dist[u][v], w); dist[v][u] = min(dist[v][u], w); // 如果是有向图,则去掉这一行 } // 执行 Floyd-Warshall 算法 floyd(); // 输出起点到终点的最短距离 cout << (dist[1][n] >= INF ? -1 : dist[1][n]) << endl; } return 0; } ``` --- #### 关键点解析 1. **邻接矩阵初始化** 使用二维数组 `dist` 存储每一对节点间的最小距离。初始状态下,设所有节点对的距离为无穷大 (`INF`),而同一节点自身的距离为零[^4]。 2. **输入处理** 对于每条边 `(u, v)` 和权重 `w`,将其存储至邻接矩阵中,并取较小值以防止重边的影响[^4]。 3. **核心逻辑** Floyd-Warshall 的核心在于三重循环:依次尝试通过中间节点优化其他两节点间的距离关系。具体而言,若从节点 \( i \) 到 \( j \) 可经由 \( k \) 达成更优解,则更新对应位置的值[^4]。 4. **边界条件** 若最终得到的结果仍为无穷大(即无法连通),则返回 `-1`;否则输出实际距离[^4]。 --- #### 性能评估 由于题目限定 \( N \leq 100 \),因此 \( O(N^3) \) 的时间复杂度完全可以接受。此外,空间需求也较低,适合此类场景下的应用。 ---
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