poj 3613Cow Relays

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 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi  ≤ 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: lengthi , I1i , and I2i

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
Source

USACO 2007 November Gold

题面

用一个矩阵a(i, j)来表示i到j经过若干条边的最短路，
初始化a为i到j边的长度，没有则是正无穷。
比如a矩阵表示经过n条边，b矩阵表示经过m条边，
那么a * b得到的矩阵表示经过m + n条边，
采用Floyd的思想进行更新。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #include<string>
 8 #include<map>
 9 #define ll long long
10 using namespace std;
11 ll n,m,S,T,l;
12 map<ll,ll>id;
13 struct node{
14     ll a[202][202];
15     friend node operator *(node x,node y)
16     {
17          node z;
18          memset(z.a,0x3f,sizeof(z.a));
19          for(ll k=1;k<=l;k++)
20           for(ll i=1;i<=l;++i)
21            for(ll j=1;j<=l;++j)
22             z.a[i][j]=min(z.a[i][j],x.a[i][k]+y.a[k][j]);
23          return z;
24     }
25 }s,ans;
26 void ksm()
27 {
28     ans=s;
29     n--;
30     while(n)
31     {
32         if(n&1) ans=ans*s;
33         s=s*s;
34         n>>=1;
35     }
36 }
37 int main()
38 {
39     freopen("run.in","r",stdin);
40     freopen("run.out","w",stdout);
41     memset(s.a,0x3f,sizeof(s.a));
42     scanf("%lld%lld%lld%lld",&n,&m,&S,&T);
43     for(ll i=1,x,y,z;i<=m;++i)
44     {
45        scanf("%lld%lld%lld",&z,&x,&y);
46        if(id[x]) x=id[x];
47        else l++,id[x]=l,x=l;
48        if(id[y]) y=id[y];
49        else l++,id[y]=l,y=l;
50        s.a[x][y]=s.a[y][x]=z;
51     }
52     S=id[S];T=id[T];
53     ksm();
54     printf("%lld",ans.a[S][T]);
55     return 0;
56 }
57 /*
58 2 6 6 4
59 11 4 6
60 4 4 8
61 8 4 9
62 6 6 8
63 2 6 9
64 3 8 9
65 10
66 */

原文地址：https://www.cnblogs.com/adelalove/p/9835410.html