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#include <algorithm>
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#include <functional>
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#include <numeric>
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#include <iostream>
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#include <iomanip>
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#include <cstdio>
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#include <cmath>
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#include <complex>
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#include <cstdlib>
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#include <ctime>
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#include <cstring>
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#include <cassert>
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#include <string>
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#include <vector>
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#include <list>
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#include <map>
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#include <set>
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#include <deque>
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#include <queue>
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#include <stack>
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#include <bitset>
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#include <sstream>
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using namespace std;
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#define LL long long
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#define LD long double
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#define PR pair<int,int>
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#define Fox(i,n) for (i=0; i<n; i++)
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#define Fox1(i,n) for (i=1; i<=n; i++)
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#define FoxI(i,a,b) for (i=a; i<=b; i++)
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#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
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#define FoxR1(i,n) for (i=n; i>0; i--)
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#define FoxRI(i,a,b) for (i=b; i>=a; i--)
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#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
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#define Min(a,b) a=min(a,b)
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#define Max(a,b) a=max(a,b)
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#define Sz(s) int((s).size())
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#define All(s) (s).begin(),(s).end()
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#define Fill(s,v) memset(s,v,sizeof(s))
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#define pb push_back
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#define mp make_pair
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#define x first
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#define y second
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template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
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template<typename T> T Sqr(T x) { return(x*x); }
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const int INF = (int)1e9;
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const LD EPS = 1e-12;
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const LD PI = acos(-1.0);
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bool Read(int &x)
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{
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char c,r=0,n=0;
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x=0;
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for(;;)
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{
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c=getchar();
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if ((c<0) && (!r))
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return(0);
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if ((c=='-') && (!r))
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n=1;
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else
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if ((c>='0') && (c<='9'))
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x=x*10+c-'0',r=1;
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else
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if (r)
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break;
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}
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if (n)
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x=-x;
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return(1);
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}
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int main()
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{
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int T,t;
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int N,M,K;
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int i,j,k,d,ans;
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int A[5000],B[5000];
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static int dist[100][100],dyn[5001][101][2];
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Read(T);
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Fox1(t,T)
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{
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Read(N),Read(M),Read(K);
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Fill(dist,60);
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Fox(i,N)
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dist[i][i]=0;
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while (M--)
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{
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Read(i),Read(j),Read(k),i--,j--;
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Min(dist[i][j],k);
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Min(dist[j][i],k);
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}
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Fox(i,N)
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Fox(j,N)
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Fox(k,N)
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Min(dist[j][k],dist[j][i]+dist[i][k]);
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Fox(i,K)
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Read(A[i]),Read(B[i]),A[i]--,B[i]--;
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Fox(i,K)
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if (max(dist[0][A[i]],dist[0][B[i]])>INF)
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{
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ans=-1;
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goto Done;
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}
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Fill(dyn,60);
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dyn[0][0][0]=0;
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Fox(i,K)
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Fox(k,2)
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Fox(j,N)
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if ((d=dyn[i][j][k])<INF)
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if (!k)
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Min(dyn[i][A[i]][1],d+dist[j][A[i]]);
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else
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{
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Min(dyn[i+1][B[i]][0],d+dist[j][B[i]]);
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if (i<K-1)
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Min(dyn[i+1][B[i]][1],d+dist[j][A[i+1]]+dist[A[i+1]][B[i]]);
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}
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ans=dyn[K][B[K-1]][0];
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Done:;
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printf("Case #%d: %d\n",t,ans);
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}
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return(0);
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} |