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 9Sample Output
10Source
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define inf 0x3f3f3f3f
#define maxn 300
int map[maxn][maxn],temp[maxn][maxn],ans[maxn][maxn];
int used[10*maxn];
int p[10*maxn];
using namespace std;
void floyd(int a[][maxn],int b[][maxn],int c[][maxn],int cnt)
{
for(int k=1;k<=cnt;k++)
for(int i=1;i<=cnt;i++)
for(int j=1;j<=cnt;j++)
if(a[i][k]+b[k][j]<c[i][j])
c[i][j]=a[i][k]+b[k][j];
}
void copy(int n,int a[][maxn],int b[][maxn])
{
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
{a[i][j]=b[i][j];
b[i][j]=inf;
}
}
int solve(int s,int t,int n,int cnt)
{
while(n)
{
if(n&1)
{
floyd(ans,map,temp,cnt);
copy(cnt,ans,temp);
}
floyd(map,map,temp,cnt);
copy(cnt,map,temp);
n>>=1;
}
return ans[s][t];
}
int main()
{int n,t,s,e;
memset(used,0,sizeof(used));
scanf("%d%d%d%d",&n,&t,&s,&e);
int u,v,w;
int cnt=0;
for(int i=0;i<=250;i++)
{for(int j=0;j<=250;j++)
{ans[i][j]=temp[i][j]=map[i][j]=inf;}
ans[i][i]=0;
}
for(int i=0;i<t;i++)
{scanf("%d%d%d",&w,&u,&v);
if(!used[u])
{
used[u]=1;
p[u]=++cnt;
}
if(!used[v])
{
used[v]=1;
p[v]=++cnt;
}
map[p[u]][p[v]]=map[p[v]][p[u]]=w;
}
printf("%d\n",solve(p[s],p[e],n,cnt));
return 0;
}