poj3613：Cow Relays（倍增优化+矩阵乘法floyd+快速幂）

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7825		Accepted: 3068

Description

For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.

Each trail connects two different intersections (1 ≤ I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the length_i of each trail (1 ≤ length_i  ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.

To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.

Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.

Input

* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: length_i , I_1i , and I_2i

Output

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

Sample Input

2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9

Sample Output

10

题意

给出一张图，求k边最短路，即经过k条边的最短路。

分析

思考一下：如果一个矩阵，表示走k条边后，一张图的点与点的最短路径，(a,b)表示从a到b的最短路径，然后我们把它与自己，按照矩阵乘法的格式“相乘”，把其中的乘改为取min，c.a[i][j] = min(c.a[i][j],x.a[i][k]+y.a[k][j]);看不懂先看下面。

这样得到的是走k+k条边的矩阵。有点抽象，下面详细解释下：

c中的一个点(a,b)，当我们用x矩阵和y矩阵求它时，我们枚举了x矩阵的a行所有数，与y矩阵的b列所有数，并且他们的坐标只能是相对应的，比如x矩阵的(a,2)这个点，相应的y矩阵点就是(2,b)，那么放到图上去理解，即从a点经过2点到b点的距离，类似的点不只有2，把所有点枚举完后，c.a[a][b]就是从a到b的最短距离。（意会一下）

这样下来，会得到走k+k条边的最短路径，对于其他的矩阵这样操作，得到的是他们两个，经过的边数相加的结果。（一个经过a条边后的矩阵 与 一个经过b条边后的矩阵这样操作后，是经过a+b条边后的矩阵，矩阵中存的是最短路径）。解释一下：向上面的例子一样，(a,2)(2,b)，是即从a点经过2点到b点的距离,因为x矩阵和y矩阵都是走k条边后的最短路径，那么x矩阵中的(a,2)是走k步后的最短路径，(2,b)也是，那么他们相加不就是走k+k条边后的最短路径吗？其他的矩阵一样。

然后，就可以套用快速幂的模板了，只不过将以前的乘改成加了，也就是倍增的思想的，比如对于走10条边，它的二进制是1010，那么我们就让在走2（10）边时的矩阵 乘以 8（1000）边的矩阵，得到走10条边的矩阵即开始时由1->2->4->8->16……即倍增中的2次幂。

code

 #include<cstdio>
 #include<algorithm>
 #include<map>
 #include<cstring>

 using namespace std;
 const int MAXN = ;

 int Hash[];
 int n,k,q,z,tn; 

 int read()
 {
     int x = ,f = ;char ch = getchar();
     while (ch<''||ch>'') {if (ch='-') f=-;ch = getchar(); }
     while (ch>=''&&ch<=''){x = x*+ch-'';ch = getchar(); }
     return x*f;
 }

 struct Matrix{
     int a[MAXN][MAXN];
     Matrix operator * (const Matrix &x) const
     {
         Matrix c;
         memset(c.a,0x3f,sizeof(c.a));
         for (int k=; k<=tn; k++)
             for (int i=; i<=tn; ++i)
                 for (int j=; j<=tn; ++j)
                     c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);
         return c;
      }
 }s,ans;
 void ksm()
 {
     ans = s;
     k--;
     for (; k; k>>=)
     {
         if (k&) ans = ans*s;
         s = s*s;
     }
 }
 int main()
 {
     k = read();n = read();q = read();z = read();
     memset(s.a,0x3f,sizeof(s.a));
     for (int x,y,w,i=; i<=n; ++i)
     {
         w = read();x = read();y = read();
         if (!Hash[x]) Hash[x] = ++tn;
         if (!Hash[y]) Hash[y] = ++tn;
         s.a[Hash[x]][Hash[y]] = s.a[Hash[y]][Hash[x]] = w;
     }
     ksm();
     printf("%d",ans.a[Hash[q]][Hash[z]]);
     return ;
 }