[vijos1046] 观光旅游

题目链接

题意：在图中找一个最小的经过三个以上结点的环

参考：http://blog.csdn.net/olga_jing/article/details/49928443

http://blog.csdn.net/zy691357966/article/details/45673647

用floyd的思想O(n3)处理出解

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int inf=2333333;
 5 const int maxn=110;
 6 const int maxm=10010;
 7 int n,m,ans;
 8 int Road[maxn][maxn];
 9 int Dist[maxn][maxn];
10 int read(){
11     int x=0,f=1;
12     char ch=getchar();
13     while (ch<‘0‘||ch>‘9‘) {
14         if (ch==‘-‘) f=-1;
15         ch=getchar();
16     }
17     while (ch>=‘0‘&&ch<=‘9‘){
18         x=x*10+ch-‘0‘;
19         ch=getchar();
20     }
21     return x*f;
22 }
23 int min(int a,int b){
24     return a>b?b:a;
25 }
26 void floyd(){
27     for (int i=1;i<=n;i++)
28       for (int j=1;j<=n;j++)
29         Dist[i][j]=Road[i][j];
30     for (int k=1;k<=n;k++){
31         for (int i=1;i<k;i++)
32             for (int j=i+1;j<k;j++)//注意循环顺序，此时前k-1个点已经处理完毕
33                 ans=min(ans,Road[i][k]+Road[k][j]+Dist[i][j]);
34         for (int i=1;i<=n;i++)
35           for (int j=1;j<=n;j++)//一般floyd求最短距离的部分
36              Dist[i][j]=min(Dist[i][j],Dist[i][k]+Dist[k][j]);
37     }
38 }
39 int main(){
40     //freopen("tour.in","r",stdin);
41     //freopen("tour.out","w",stdout);
42     while (~scanf("%d%d",&n,&m)){
43         for (int i=0;i<=n;i++){
44           for (int j=0;j<=n;j++)
45             Road[i][j]=inf;
46           Road[i][i]=0;
47         }
48         ans=inf;
49         for (int i=0;i<m;i++){
50             int u=read(),v=read(),w=read();
51             Road[u][v]=Road[v][u]=min(Road[u][v],w);
52         }
53         floyd();
54         if (ans<inf) printf("%d\n",ans);
55         else printf("No solution.\n");
56     }
57     return 0;
58 }