## Solución al problema número HISIX de Spoj - HISIX.

Cualquier duda no dudes en contactar.

``````#include <bits/stdc++.h>
#define INF 0x3F3F3F3F
using namespace std;

const int MAXN = 200002;

struct edge{
int from, to, weight;
edge(){}
edge(int a, int b, int c){
from = a;
to = b;
weight = c;
}
};

struct state{
int node, dist;
state(){}
state(int a, int b){
node = a; dist = b;
}
bool operator<(const state &other)const{ // sobrecarga de operadores para ordenar
return other.dist < dist;
}
};

vector<edge> graph[MAXN];
int dist[MAXN];
int N,E;

int dijkstra(int start, int end){
dist[start] = 0;
priority_queue<state> pq;
pq.push(state(start, 0));
while(!pq.empty()){
state cur = pq.top(); pq.pop();
if(dist[cur.node] < cur.dist) continue;
if(cur.node == end) return cur.dist;
for(unsigned int i=0;i<graph[cur.node].size();i++){
int dest = graph[cur.node][i].to;
int wht = graph[cur.node][i].weight + cur.dist;
if(dist[dest] <= wht) continue;
dist[dest] = wht;
pq.push(state(dest, wht));
}
}
return -1;
}
int main(){
int cases,c=1; scanf("%d",&cases);
while(cases--)
{
scanf("%d",&N);
for(int i=0;i<=N*2;i++) { dist[i]=INF; graph[i].clear(); }
map<string,int> M;
int count=1;
for(int i=0; i<N;i++)
{
string n1,n2; int weight; cin>>n1>>n2>>weight;
if(M[n1]==0) {M[n1]=count; count++;}
if(M[n2]==0) {M[n2]=count; count++;}
graph[M[n1]].push_back(edge(M[n1],M[n2],weight));
graph[M[n2]].push_back(edge(M[n2],M[n1],weight));
}
int T=-INF;
for(int i=1; i<count;i++)
{
dijkstra(i,count);
for(int i=1;i<count;i++) { T=max(T,dist[i]); dist[i]=INF;}
}
if(T==INF) printf("Case #%d: INFINITE\n",c++);
else printf("Case #%d: %d\n",c++,T);
}
return 0;
}
``````