There is a travel agency in Adelton town on Zanzibar island. It has decided to
offer its clients, besides many other attractions, sightseeing the town. To
earn as much as possible from this attraction, the agency has accepted a
shrewd decision: it is necessary to find the shortest route which begins and
ends at the same place.
Your task is to write a program which finds such a route.
In the town there are N crossing points numbered from 1 to N and M two-way
roads numbered from 1 to M. Two crossing points can be connected by multiple
roads, but no road connects a crossing point with itself. Each sightseeing
route is a sequence of road numbers y1, …, yk, k > 2. The road yi
(1 ≤ i ≤ k ? 1) connects crossing points xi and xi+1, the road yk connects
crossing points xk and x1. All the numbers x1, …, xk should be different.
The length of the sightseeing route is the sum of the lengths of all roads on
the sightseeing route, i.e. L(y1) + L(y2) + … + L(yk) where L(yi) is the
length of the road yi (1 ≤ i ≤ k). Your program has to find such a sightseeing
route, the length of which is minimal, or to specify that it is not possible,
because there is no sightseeing route in the town.
Input
Input contains T tests (1 ≤ T ≤ 5).
The first line of each test contains two integers: the
number of crossing points N and the number of roads M (3 ≤ N ≤ 100; 3 ≤ M ≤ N · (N ? 1)).
Each of the next M lines describes one road. It contains 3 integers: the number
of its first crossing point a, the number of the second one b, and the length of the road l (1 ≤ a, b ≤ N; a ≠ b; 1 ≤ l ≤ 300). Input is ended with a “?1” line.
Output
Each line of output is an answer. It contains either a string
“No solution.” in case there isn‘t any sightseeing route, or it contains the
numbers of all crossing points on the shortest sightseeing route in the order
how to pass them (i.e. the numbers x1 to xk from our definition of a
sightseeing route), separated by single spaces. If there are multiple
sightseeing routes of the minimal length, you can output any one of them.
Sample
| input | output |
|---|
5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
4 3
1 2 10
1 3 20
1 4 30
-1
|
1 3 5 2
No solution.
|
Problem Source: Central European Olympiad in Informatics 1999
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 1e2+10;
const int M = 24005;
const int INF=0x7ffffff;
int dist[N][N], w[N][N];
int pre[N][N];
int path[N];
int n, m, num, minc;
void floyd() {
minc=INF;
for(int k=1; k<=n; k++) {
for(int i=1; i<k; i++)
for(int j=i+1; j<k; j++) {
int ans=dist[i][j]+w[i][k]+w[k][j];
if(ans<minc) { //ж‰?е€°ж??д??и§?
minc=ans;
num=0;
int p=j;
while(p!=i) { //й??е?‘е??ж‰?е‰?й?±й??е??з??и·?е??е№?е°?е…?е-?е?Ёиμ·ж??
path[num++]=p;
p=pre[i][p];
}
path[num++]=i;
path[num++]=k;
}
}
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++) {
if(dist[i][j]>dist[i][k]+dist[k][j]) {
dist[i][j]=dist[i][k]+dist[k][j];
pre[i][j]=pre[k][j];
}
}
}
}
int main() {
int u, v, cost;
while(cin >> n) {
if(n<0) break;
cin >> m;
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++) {
dist[i][j]=w[i][j]=INF;
pre[i][j]=i;
}
for(int i=1; i<=m; i++) {
scanf("%d%d%d",&u,&v,&cost);
if(dist[u][v]>cost) //е¤?з??й??и?№
w[u][v]=w[v][u]=dist[u][v]=dist[v][u]=cost;
}
floyd();
if(minc==INF)
printf("No solution.\n");
else {
printf("%d",path[0]);
for(int i=1; i<num; i++)
printf(" %d",path[i]);
puts("");
}
}
return 0;
}
