[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

#include<cstdio>
#include<cstring>
using namespace std;
const int inf=2333333;
const int maxn=110;
const int maxm=10010;
int n,m,ans;
int Road[maxn][maxn];
int Dist[maxn][maxn];
int read(){
    int x=0,f=1;
    char ch=getchar();
    while (ch<'0'||ch>'9') {
        if (ch=='-') f=-1;
        ch=getchar();
    }
    while (ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
int min(int a,int b){
    return a>b?b:a;
}
void floyd(){
    for (int i=1;i<=n;i++)
      for (int j=1;j<=n;j++)
        Dist[i][j]=Road[i][j];
    for (int k=1;k<=n;k++){
        for (int i=1;i<k;i++)
            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
                ans=min(ans,Road[i][k]+Road[k][j]+Dist[i][j]);
        for (int i=1;i<=n;i++)
          for (int j=1;j<=n;j++)//一般floyd求最短距离的部分
             Dist[i][j]=min(Dist[i][j],Dist[i][k]+Dist[k][j]);
    }
}
int main(){
    //freopen("tour.in","r",stdin);
    //freopen("tour.out","w",stdout);
    while (~scanf("%d%d",&n,&m)){
        for (int i=0;i<=n;i++){
          for (int j=0;j<=n;j++)
            Road[i][j]=inf;            
          Road[i][i]=0;
        }
        ans=inf;
        for (int i=0;i<m;i++){
            int u=read(),v=read(),w=read();
            Road[u][v]=Road[v][u]=min(Road[u][v],w);
        }
        floyd();
        if (ans<inf) printf("%d
",ans);
        else printf("No solution.
");
    }
    return 0;
}