Sightseeing trip
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 5038
Accepted: 1930
Special Judge
Description
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
y_1, ..., 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 x_1. All the numbers x_1,...,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(y_1)+L(y_2)+...+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
The first line of input contains two positive integers: the number of crossing points N<=100 and the number of roads M<=10000. Each of the next M lines describes one road. It contains 3 positive integers:
the number of its first crossing point, the number of the second one, and the length of the road (a positive integer less than 500).
Output
There is only one line in output. 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 x_1 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
5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
Sample Output
1 3 5 2
Source
CEOI 1999
题目大意：有个旅游公司要开发一条新的旅游路线，要求这条路尽可能短，但是又不能只包含
两个城市，并且旅游途中不能回到之前去过的城市，只能去往下一个没去过的城市，旅游结束
的时候要回到最开始的城市，要求求出整个旅游路线经过的城市。
思路：给N个点，M条边建图。路程最短，且要形成环，其实就是求最小环问题。可以用Floyd
来做。用Dist[i][j]存储从i到j的最短路径，但是 i != j，因为最少要有3个点(加上k至少3个点)。
用pre[i][j]来表示从点i到点j的路径中j点前边的点。判定最小环时，点i到j的路径再加上点k就是
当前的最小环，利用pre[i][j]从点j反向找到点i，在加上k就是最小环了。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 110;
const int INF = 0xffffff0;
int temp,Map[MAXN][MAXN],Dist[MAXN][MAXN],pre[MAXN][MAXN],ans[MAXN*3];
void Solve(int i,int j,int k)
{
temp = 0; //回溯，存储最小环
while(i != j)
{
ans[temp++] = j;
j = pre[i][j];
}
ans[temp++] = i;
ans[temp++] = k;
}
void Floyd(int N)
{
for(int i = 1; i <= N; ++i)
for(int j = 1; j <= N; ++j)
{
Dist[i][j] = Map[i][j];
pre[i][j] = i;
}
int MinCircle = INF;
for(int k = 1; k <= N; ++k)
{
for(int i = 1; i <= N; ++i)
{
for(int j = 1; j <= N; ++j)
{
if(i != j && Dist[i][j] != INF && Map[i][k] != INF && Map[k][j] != INF
&& Dist[i][j] + Map[i][k] + Map[k][j] < MinCircle)
{
MinCircle = min(MinCircle, Dist[i][j] + Map[i][k] + Map[k][j]);
Solve(i,j,k);
}
}
}
for(int i = 1; i <= N; ++i)
{
for(int j = 1; j <= N; ++j)
{
if(Dist[i][k] != INF && Dist[k][j] != INF &&
Dist[i][k] + Dist[k][j] < Dist[i][j])
{
Dist[i][j] = Dist[i][k] +Dist[k][j];
pre[i][j] = pre[k][j]; //记录点i到点j的路径上，j前边的点
//pre[j][i] = pre[k][i];
}
}
}
}
if(MinCircle == INF)
{
printf("No solution.\n");
return;
}
for(int i = 0;i < temp; ++i)
if(i != temp-1)
printf("%d ",ans[i]);
else
printf("%d\n",ans[i]);
}
int main()
{
int N,M,u,v,w;
while(~scanf("%d%d",&N,&M))
{
for(int i = 1; i <= N; ++i)
for(int j = 1; j <= N; ++j)
Map[i][j] = INF;
for(int i = 0; i < M; ++i)
{
scanf("%d%d%d",&u,&v,&w);
if(w < Map[u][v])
Map[u][v] = Map[v][u] = w;
}
Floyd(N);
}
return 0;
}