poj 1734 Sightseeing trip判断最短长度的环

Sightseeing trip

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5590		Accepted: 2151		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

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网上的题解都是用floyd做出来的，用floyd最短路算法判断是否有环，但是我感觉dfs足可以解决这道题

下面的是我的代码和题解，后面我还会贴上网上大牛用floyd写的算法

dfs：以边为核心，边上的两个点一个为起点，一个为终点，开始进行dfs搜索。每次从起点出发，如果找到终点并且路程变小就记录下路径，在后输出最短路的路径即可

如过ans的值和起始值一样的话，就相当于没有环出现，那么输出那个字符串就可以了

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf=;
int n,m;
int Map[][];
int head[];
int temp;
bool vis[];
int a[];
int start ,end;
int ans;
bool judge;
int t;
int cnt;
int ans_num;
void dfs(int u,int node,int dis){
 
     if(u==end){
              if(dis<ans){
                  ans=dis;
 
            for(int i=;i<node;i++)
                   a[i]=head[i];
            a[node]=u;
            ans_num=node;
              }
              return ;
     }
    head[node]=u;
      for(int i=;i<=n;i++){
           if(!Map[u][i]||vis[i])
               continue;
          vis[i]=true;
          if(dis+Map[u][i]<ans)
              dfs(i,node+,dis+Map[u][i]);
          vis[i]=false;
      }
      return;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        if(n==-)
            break;
        memset(Map,,sizeof(Map));
        scanf("%d",&m);
        int u,v,w;
        for(int i=;i<m;i++){
           scanf("%d%d%d",&u,&v,&w);
 
           if(!Map[u][v]){
                Map[u][v]=w;
                Map[v][u]=w;
           }
           else{
              if(w<Map[u][v]){
                  Map[u][v]=w;
                  Map[v][u]=w;
              }
           }
        }
             ans=inf;
        for(int i=;i<=n;i++)
            for(int j=i+;j<=n;j++){
                if(Map[i][j]==)
                continue;
            memset(vis,false,sizeof(vis));
            t=Map[i][j];
               Map[i][j]=;
               Map[j][i]=;
             start=i;
              end=j;
           vis[i]=true;
           dfs(start,,t);        
 
            Map[i][j]=t;
            Map[j][i]=t;
        }
        if(ans!=inf){
 
            for(int  i=;i<=ans_num;i++)
                printf("%d%c",a[i],i==ans_num?'\n':' ');
        }
 
          else
 
              printf("No solution.\n");
    }
    return ;
}

floyd算法代码

给出一张无向图，求一个最小环并输出路径。

说说我的感觉：

包含点 i 和点 j 的最小环，我们可以看成是 i 到 j 之间的最短路和次短路的组合，通过 floyd 可求任意两点之间的最短距离，那么我们只要找到最短路径外的一条最短路来保证 i 和 j 之间可达即可。在做 floyd 循环的同时，我们以 环权值 最小（最短路权值+次短路权值=最小环权值）为标准，一直更新每个点的前驱，也就是记录 i 到 j 的最短路径，以及，能够松弛 i 和 j 的点 k （k 不在 i 到 j 的最短路径中）中代价最小的那个（也就是 i 到 j 之间的次短路），然后按环的自然顺序输出即可。

代码中也注释的很详细了：

#include<cstdio>
#include<cstring>
#define find_min(a,b) a<b?a:b
 
const int N = ;
const int INF = 0x7ffffff;
int mat[N][N],dist[N][N],pre[N][N],path[N],n;
 
int main()
{
    int i,j,k,m,a,b,c;
    int num;
 
    while(~scanf("%d%d",&n,&m)){
        for(i=;i<=n;i++){
            for(j=;j<=n;j++){
                mat[i][j]=dist[i][j]=INF;
                pre[i][j]=i;
            }
        }
        while(m--){
            scanf("%d%d%d",&a,&b,&c);
            mat[a][b]=mat[b][a]=dist[a][b]=dist[b][a]=find_min(mat[a][b],c);
        }
 
        int min=INF;
        for(k=;k<=n;k++){//最短路径外一点将最短路首尾链接,那么就得到一个最小环
            for(i=;i<k;i++){
                for(j=i+;j<k;j++){
                    //求最小环不能用两点间最短路松弛,因为(i,k)之间的最短路,(k,j)之间的最短路可能有重合的部分
                    //所以mat[][]其实是不更新的,这里和单纯的floyd最短路不一样
                    //dist[i][j]保存的是 i 到 j 的最短路权值和
                    int tmp=dist[i][j]+mat[i][k]+mat[k][j];//这里 k 分别和 i 还有 j 在mat中直接相连
                    if(tmp<min){
                        min=tmp;
                        num=;
                        int p=j;
                        while(p!=i){//回溯
                            path[num++]=p;
                            p=pre[i][p];//pre[i][j]表示 i 到 j 最短路径上 j 前面的一个点
                        }
                        path[num++]=i;
                        path[num++]=k;
                    }
                }
            }
            for(i=;i<=n;i++){
                for(j=;j<=n;j++){
                    if(dist[i][j]>dist[i][k]+dist[k][j]){
                        dist[i][j]=dist[i][k]+dist[k][j];//dist[][]保存两点间最短距离
                        pre[i][j]=pre[k][j];
                    }
                }
            }
        }
        if(min==INF)puts("No solution.");
        else{
            printf("%d",path[]);
            for(i=;i<num;i++)
                printf(" %d",path[i]);
            puts("");
        }
    }
    return ;
}