his journey to the west. Liyuan thought it is too brutal for the monkey, so he changed the story:
One day, Wukong left his home - Mountain of Flower and Fruit, to the Dragon King’s party, at the same time, Tang Monk left Baima Temple to the Lingyin Temple to deliver a lecture. They are both busy, so they will choose the shortest path. However, there may
be several different shortest paths between two places. Now the Buddha wants them to encounter on the road. To increase the possibility of their meeting, the Buddha wants to arrange the two routes to make their common places as many as possible. Of course,
the two routines should still be the shortest paths.
Unfortunately, the Buddha is not good at algorithm, so he ask you for help.
integers a b c, indicating there is a road between place a and b, whose length is c. Please note the roads are undirected. The last line contains four integers A B C D, separated by spaces, indicating the start and end points of Wukong, and the start and end
points of Tang Monk respectively.
The input are ended with N=M=0, which should not be processed.
6 6 1 2 1 2 3 1 3 4 1 4 5 1 1 5 2 4 6 3 1 6 2 4 0 0
3 Hint: One possible arrangement is (1-2-3-4-6) for Wukong and (2-3-4) for Tang Monk. The number of common points are 3.
Floyd的应用
用Floyd求出任意两点间的最短路径
两个定理：
1 .所求的路径一定是一断连续的路径
2.如果路径(x,y)是a->b的最短路径中的一段，则min(a,b) = min(a,x) + min(x,y) + min(y,b) 最后只需找到同时在两条最短路径上，且距离最长的那一段
代码：
#include<iostream>
using namespace std;
#define INF 200000000
int map[305][305],M[350][305],A[305][305];
int n,m;
void Floyd()
{
int i,j,k;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
A[i][j]=map[i][j];
}
for(k=1;k<=n;k++)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(A[i][j]>A[i][k]+A[k][j])
{
A[i][j]=A[i][k]+A[k][j];
M[i][j]=M[i][k]+M[k][j];
}
else if(A[i][j]==A[i][k]+A[k][j]&&M[i][j]<M[i][k]+M[k][j])
{
M[i][j]=M[i][k]+M[k][j];
}
}
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m),m||n)
{
if(n==0&&m==0)
break;
memset(M,0,sizeof(M));
for(i=0;i<=n;i++)
{
for(j=0;j<=n;j++)
map[i][j]=(i==j)?0:INF;
}
int a,b,w;
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&w);
if(map[a][b]<w)
continue;
map[a][b]=w;
map[b][a]=w;
M[a][b]=1;
M[b][a]=1;
}
Floyd();
int A1,B,C,D;
scanf("%d%d%d%d",&A1,&B,&C,&D);
int ans=-1;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(M[i][j]>ans&&(A[A1][B]==A[A1][i]+A[i][j]+A[j][B])&&(A[C][D]==A[C][i]+A[i][j]+A[j][D]))
ans=M[i][j];
}
}
printf("%d\n",ans+1);
}
return 0;
}