C++ :

/*其实这个题就是把点当成面来给了，对于某一个区域，把他当成一个点，
和他相邻的区域与他之间有一条权为1的边，
*/
#include<stdio.h>
#include<string.h>
#define Max 210
#define INF 10000000
#define M 210
#define N 260

bool map[M][M];

struct Area
{
	int town;    //在该区域边上城镇数
	int turn[N]; //这些城镇依次是turn[n];
	int f[N];    //f[i]记i是turn[n]的第f[i]个数
}area[M];

bool check(int x, int y)
{
	int i, a, b, t;
	bool yes = 0;
	for(i = 0; i < area[y].town; i++){
		a = area[y].turn[i]; 
		b = area[y].turn[(i+1)%area[y].town];		
		if(area[x].f[a] != -1 && area[x].f[b] != -1)
		{
			t = area[x].f[a]-area[x].f[b];
			if(t == -1 || t == 1 || t == area[x].town-1 || t == 1-area[x].town)
			{	yes = 1; break; }
		}
	}
	return yes;
}

int A[Max][Max];
void Floyd(int n)
{
	int i, j, k;
	for(i = 0; i < n; i++){
		for(j = 0; j < n; j++){
			if(i != j && !map[i][j])
				A[i][j] = INF;
			else A[i][j] = map[i][j];
		}
	}
	for(k = 0; k < n; k++){
		for(i = 0; i < n; i++)
			for(j = 0; j < n; j++)
				if(A[i][j] > A[i][k]+A[k][j])
					A[i][j] = A[i][k]+A[k][j];
	}
}

int main()
{
	int i, j, m, n, l, man[40];
	int town[N][N], num[N];
		//freopen("walls.in", "r", stdin);
	//freopen("walls.out", "w", stdout);
	scanf("%d", &m);
		scanf("%d", &n);
		scanf("%d", &l);
		for(i = 0; i < l; i++)
			scanf("%d", &man[i]);

		memset(map, 0, sizeof(map));
		memset(num, 0, sizeof(num));
		for(i = 0; i < m; i++){
			scanf("%d", &area[i].town);	
			memset(area[i].f, -1, sizeof(area[i].f));
			for(j = 0; j < area[i].town; j++){
				scanf("%d", &area[i].turn[j]);
				area[i].f[area[i].turn[j]] = j;
				town[area[i].turn[j]][num[area[i].turn[j]]] = i;
				num[area[i].turn[j]]++;
			}
			for(j = 0; j < i; j++)
				map[i][j] = map[j][i] = check(i, j);
		}
		Floyd(m);
		int minsum, sum, min, k;
		for(i = 0; i < m; i++){
			sum = 0;
			for(j = 0; j < l; j++){
				min = A[town[man[j]][0]][i];
				for(k = 1; k < num[man[j]]; k++){
					if(min > A[town[man[j]][k]][i])
						min = A[town[man[j]][k]][i];
				}
				sum += min;
			}
			if(!i) minsum = sum;
			else if(minsum > sum)
				minsum = sum;
		}
		printf("%d\n", minsum);

//	}
	return 0;
}

Pascal :

var an,i,j,k,l,n,m,x,y,z:longint;
a,b:array[0..200,0..200]of longint;
c:array[1..30]of longint;
d:array[0..250,0..201]of longint;
procedure qzd;
begin
z:=maxlongint;
for i:=1 to m do
begin
x:=0;
for j:=1 to l do
begin
y:=maxlongint;
for k:=1 to d[c[j],0] do
if a[d[c[j],k],i]<y then y:=a[d[c[j],k],i];
inc(x,y);
end;
if x<z then z:=x;
end;
write(z);
end;
procedure hjm;
begin
for k:=1 to m do
for i:=1 to m do
for j:=1 to m do
if a[i,k]+a[k,j]<a[i,j] then a[i,j]:=a[i,k]+a[k,j];
end;
procedure cz;
begin
readln(m);
readln(n);
readln(l);
for i:=1 to l do
read(c[i]);
readln;
for i:=0 to m do
for j:=0 to m do
a[i,j]:=1000;
for i:=1 to m do
begin
readln(k);
read(x,y);
z:=x;
if b[y,x]>0 then begin
a[i,b[y,x]]:=1;
a[b[y,x],i]:=1;
end
else b[x,y]:=i;
inc(d[x,0]);
d[x,d[x,0]]:=i;
inc(d[y,0]);
d[y,d[y,0]]:=i;
for j:=3 to k do
begin
x:=y;
read(y);
if b[y,x]>0 then begin
a[i,b[y,x]]:=1;
a[b[y,x],i]:=1;
end
else b[x,y]:=i;
inc(d[y,0]);
d[y,d[y,0]]:=i;
end;
x:=y;
y:=z;
if b[y,x]>0 then begin
a[i,b[y,x]]:=1;
a[b[y,x],i]:=1;
end
else b[x,y]:=i;
end;
for i:=0 to m do
a[i,i]:=0;
end;
begin
cz;
hjm;
qzd;
end.