phy rails 模板 矩陣 structure ssi 進制 size and

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

題意

給出一張圖，求k邊最短路，即經過k條邊的最短路。

分析

思考一下：如果一個矩陣，表示走k條邊後，一張圖的點與點的最短路徑，(a,b)表示從a到b的最短路徑，然後我們把它與自己，按照矩陣乘法的格式“相乘”，把其中的乘改為取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。

這樣得到的是走k+k條邊的矩陣。有點抽象，下面詳細解釋下：

c中的一個點(a,b)，當我們用a矩陣和b矩陣求它時，我們枚舉了x矩陣的a行所有數，與y矩陣的b列所有數，並且他們的坐標只能是相對應的，比如x矩陣的(a,2)這個點，相應的y矩陣點就是(2,b)，那麽放到圖上去理解，即從a點經過2點到b點的距離，類似的點不只有2，把所有點枚舉完後，c.a[a][b]就是從a到b的最短距離。（意會一下）

這樣下來，會得到走k+k條邊的最短路徑，對於其他的矩陣這樣操作，得到的是他們兩個，經過的邊數相加的結果。（一個經過a條邊後的矩陣 與 一個經過b條邊後的矩陣這樣操作後，是經過a+b條邊後的矩陣，矩陣中存的是最短路徑）。解釋一下：向上面的例子一樣，(a,2)(2,b)，是即從a點經過2點到b點的距離,因為x矩陣和y矩陣都是走k條邊後的最短路徑，那麽x矩陣中的(a,2)是走k步後的最短路徑，(2,b)也是，那麽他們相加不就是走k+k條邊後的最短路徑嗎？其他的矩陣一樣。

然後，就可以套用快速冪的模板了，只不過將以前的乘改成加了，也就是倍增的思想的，比如對於走10條邊，它的二進制是1010，那麽我們就讓在走2（10）邊時的矩陣 乘以 8（1000）邊的矩陣，得到走10條邊的矩陣即開始時由1->2->4->8->16……即倍增中的2次冪。

code

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<map>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 const int MAXN = 210;
 8 
 9 int Hash[1010];
10 int n,k,q,z,tn; 
11 
12 int read()
13 {
14     int x = 0,f = 1;char ch = getchar();
15     while (ch<‘0‘||ch>‘9‘) {if (ch=‘-‘) f=-1;ch = getchar(); }
16     while (ch>=‘0‘&&ch<=‘9‘){x = x*10+ch-‘0‘;ch = getchar(); }
17     return x*f;
18 }
19 
20 struct Matrix{
21     int a[MAXN][MAXN];
22     Matrix operator * (const Matrix &x) const 
23     {
24         Matrix c;
25         memset(c.a,0x3f,sizeof(c.a));
26         for (int k=1; k<=tn; k++)
27             for (int i=1; i<=tn; ++i)
28                 for (int j=1; j<=tn; ++j)
29                     c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);
30         return c;
31      } 
32 }s,ans;
33 void ksm()
34 { 
35     ans = s;
36     k--;
37     for (; k; k>>=1)
38     {
39         if (k&1) ans = ans*s;
40         s = s*s;
41     }
42 }
43 int main()
44 {
45     k = read();n = read();q = read();z = read();
46     memset(s.a,0x3f,sizeof(s.a));
47     for (int x,y,w,i=1; i<=n; ++i)
48     {
49         w = read();x = read();y = read();
50         if (!Hash[x]) Hash[x] = ++tn;
51         if (!Hash[y]) Hash[y] = ++tn;
52         s.a[Hash[x]][Hash[y]] = s.a[Hash[y]][Hash[x]] = w;
53     }
54     ksm();
55     printf("%d",ans.a[Hash[q]][Hash[z]]);
56     return 0;
57 }

poj3613：Cow Relays（倍增優化+矩陣乘法floyd+快速冪）