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
6 6 4
4 6
4 8
4 9
6 8
6 9
8 9
Sample Output
Source
USACO 2007 November Gold
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。
代码
include
include
include
include
include
include
include
include
define ll long long
using namespace std;
ll n,m,S,T,l;
map<ll,ll>id;
struct node{
ll a[202][202];
friend node operator (node x,node y)
{
node z;
memset(z.a,0x3f,sizeof(z.a));
for(ll k=1;k<=l;k++)
for(ll i=1;i<=l;++i)
for(ll j=1;j<=l;++j)
z.a[i][j]=min(z.a[i][j],x.a[i][k]+y.a[k][j]);
return z;
}
}s,ans;
void ksm()
{
ans=s;
n--;
while(n)
{
if(n&1) ans=anss;
s=s*s;
n>>=1;
}
}
int main()
{
freopen("run.in","r",stdin);
freopen("run.out","w",stdout);
memset(s.a,0x3f,sizeof(s.a));
scanf("%lld%lld%lld%lld",&n,&m,&S,&T);
for(ll i=1,x,y,z;i<=m;++i)
{
scanf("%lld%lld%lld",&z,&x,&y);
if(id[x]) x=id[x];
else l++,id[x]=l,x=l;
if(id[y]) y=id[y];
else l++,id[y]=l,y=l;
s.a[x][y]=s.a[y][x]=z;
}
S=id[S];T=id[T];
ksm();
printf("%lld",ans.a[S][T]);
return 0;
}