There are many pits in this question.
1. The input may be G1 G1 10. Need to filter out
2. The input may have multiple paths such as G1 G2 10, G1 G2 1. The minimum value should be kept.
3. The gas station number may reach G10 three letters, and the residential building number may reach 1000 four letters. If you look at the sample, you will mistake it for only one number.
Code:
#include <cstdio>
#include <climits>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxv = 1010;
const int INF = INT_MAX;
int N, M, V, K, Ds;
int graph[maxv][maxv];
int d[maxv];
bool confirmed[maxv];
int f_index=-1;
double f_min=0.0, f_ave=INF;
int change(char a[]);
void dijkstra(int s);
int main(){
scanf("%d%d%d%d", &N, &M, &K, &Ds);
V = N + M;
fill(graph[0], graph[0]+maxv*maxv, INF);
while(K--){
char p1[5], p2[5];
int dist;
scanf("%s%s%d", p1, p2, &dist);
int v1 = change(p1);
int v2 = change(p2);
if(v1==v2) continue;
if(dist<graph[v1][v2]){
graph[v1][v2] = graph[v2][v1] = dist;
}
}
for(int i=0; i<M; i++){
dijkstra(i);
double sum=0.0, t_ave, t_min=INF;
bool solution = true;
for(int j=M; j<V; j++){
if(d[j]>Ds){
solution = false;
break;
}
sum += d[j];
if(d[j]<t_min) t_min = d[j];
}
if(!solution) continue;
else{
t_ave = sum / N;
if(t_min>f_min || (t_min==f_min && t_ave<f_ave)){
f_min = t_min;
f_ave = t_ave;
f_index = i;
}
}
}
if(f_index==-1) printf("No Solution");
else printf("G%d\n%.1f %.1f", f_index+1, f_min, round(f_ave*10)/10);
return 0;
}
int change(char a[]){
if(a[0]=='G'){
if(strlen(a)==3) return 9;
else return a[1]-'1'; //加油站编号 0 ~ M-1
}
else{
int n;
sscanf(a, "%d", &n);
return n-1+M; //居民楼编号 M ~ V-1
}
}
void dijkstra(int s){
fill(d, d+V, INF);
fill(confirmed, confirmed+V, 0);
d[s] = 0;
for(int k=0; k<V; k++){
int u=-1, min_d=INF;
for(int i=0; i<V; i++){
if(!confirmed[i] && d[i]<min_d){
u = i;
min_d = d[i];
}
}
if(u==-1) return;
confirmed[u] = true;
for(int j=0; j<V; j++){
//u->j
if(!confirmed[j] && graph[u][j]!=INF){
if(d[u]+graph[u][j]<d[j]){
d[j] = d[u] + graph[u][j];
}
}
}
}
return;
}