URAL 1004 Sightseeing Trip（最小环）

最新推荐文章于 2019-08-20 16:25:54 发布

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最新推荐文章于 2019-08-20 16:25:54 发布

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原文链接：http://www.cnblogs.com/jianrenfang/p/6021351.html

本文介绍了一种寻找从同一地点出发并返回的最短环游路径算法，该算法适用于旅游景点路线规划等问题。通过Floyd算法求解，实现对多条路径的长度计算与比较。

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Sightseeing Trip

Time limit: 0.5 second
Memory limit: 64 MB

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, …, y_k, k > 2. The road y_i (1 ≤ i ≤ k − 1) connects crossing points x_i and x_i+1, the road y_k connects crossing points x_k and x₁. All the numbers x₁, …, x_k 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(y_k) where L(y_i) is the length of the road y_i (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 x_k 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‰ѕе€°жњЂдји§Ј
                    minc=ans;
                    num=0;
                    int p=j;
                    while(p!=i) { //йЂ†еђ‘еЇ»ж‰ѕе‰Ќй©±йЃЌеЋ†зљ„и·Їеѕ„е№¶е°†е…¶ее‚Ёиµ·жќҐ
                        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¤„зђ†й‡Ќиѕ№
                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;
}