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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 75 3 20
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;
}
POJ1734 Sightseeing trip【Floyd】【最小环】
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原文地址：http://blog.csdn.net/lianai911/article/details/43238477
