POJ 3613Cow Relays

Source: Internet
Author: User

Description

For Their physical fitness program, N (2≤n≤1,000,000) cows has decided to run a relay race using the T (2 ≤T≤100) cow trails throughout the pasture. Each trail connects-different intersections (1≤i1i≤1,000; 1≤i2i≤1,000), each of the which are the termination for a t least and trails. The cows know the lengthi of each trail (1≤lengthi≤1,000), the intersections of the trail connects, and they know th At no, intersections is directly connected by the 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 has more than one cow ). They must position themselves properly so this they can hand off the baton Cow-by-cow and end up at the proper finishing P Lace. 

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's the shortest distance from intersection S to int Ersection E that traverses exactly N cow trails. 

Sample Input

2 6 6 411 4 4 4 6 6 8 9

Sample Output

10

source

usaco November Gold

surface


A matrix A (I, j) is used to denote the shortest path from I to J through several edges,
Initialize a to the length of the I to J side, not the positive infinity.
For example, a matrix is represented by an n-bar, B-matrix is represented by M-Bar,
Then A * b gets a matrix representing the M + N edges,
The idea of using Floyd is updated.

 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 */
View Code





POJ 3613Cow Relays


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