Uva 12544 - Beehives 找无向图的最小环..BFS..

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题意:

给了一个无向图(至多700个点,两点间无重边),问其中边数最少的环是所少条边

题解:

这类问题可以用Floyd做..但是会超时...

用BFS的方法,思路就是形成了环,则必然是搜索树上有了前向或者平行边..枚举每个点位根..做BFS..按照遍历的顺序给每个点标号..当找到一个已经标号的边..则知道形成了环..距离为dis[u]+dis[v]-1...但是这种方法只能求这种边权值都为1的最小环...加些条件就很容易出错了..而Floyd的方法适用范围更广...

Program(Floyd,TLE):

#include<iostream>#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#include<cmath>#define oo 1000000007#define MAXN 505using namespace std;int Dist[MAXN][MAXN],Graph[MAXN][MAXN];int MC(int nVertex){      int mincircle=oo,i,j,k,temp;      for (i=0;i<nVertex;i++)        for (j=0;j<nVertex;j++)           Dist[i][j]=Graph[i][j];      for(k=0;k<nVertex;++k)      {          //新增部分:           for(i=0;i<k;++i)               for(j=0;j<i;++j)                  if (mincircle>Dist[i][j]+Graph[j][k]+Graph[k][i])                     mincircle = Dist[i][j]+Graph[j][k]+Graph[k][i];         //通常的 floyd 部分:           for(i=0;i<nVertex;++i)              for(j=0;j<i;++j)              {                     temp = Dist[i][k] + Dist[k][j];                     if(temp < Dist[i][j])                        Dist[i][j] = Dist[j][i] = temp;              }      }      return mincircle;}int main(){       int C,cases,N,M,u,v,ans;        scanf("%d",&C);       for (cases=1;cases<=C;cases++)       {                scanf("%d%d",&N,&M);                for (u=0;u<N;u++)                   for (v=0;v<N;v++)                      Graph[u][v]=oo;                while (M--)                {                        scanf("%d%d",&u,&v);                        Graph[u][v]=Graph[v][u]=1;                }                ans=MC(N);                printf("Case %d: ",cases);                if (ans==oo) puts("impossible");                        else printf("%d\n",ans);       }       return 0;}

Program(BFS,AC):

#include<iostream>#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#include<cmath>#define oo 1000000007#define MAXN 505using namespace std;struct node{       int v,next,id;}edge[50005];int Ne,_next[MAXN],dfn[MAXN],id[MAXN],ans; queue<int> Q;void addedge(int u,int v,int id){       edge[++Ne].next=_next[u],_next[u]=Ne;       edge[Ne].v=v,edge[Ne].id=id;} int main(){       int C,cases,N,M,u,v,i,k;         scanf("%d",&C);       for (cases=1;cases<=C;cases++)       {                scanf("%d%d",&N,&M);                memset(_next,0,sizeof(_next)),Ne=0;                for (i=1;i<=M;i++)                {                       scanf("%d%d",&u,&v),u++,v++;                       addedge(u,v,i),addedge(v,u,i);                }                ans=oo;                   for (i=1;i<=N;i++)                {                        memset(dfn,0x7f,sizeof(dfn));                        memset(id,0,sizeof(id));                                Q.push(i);                        dfn[i]=0;                        while (Q.size())                        {                               u=Q.front(),Q.pop();                               for (k=_next[u];k;k=edge[k].next)                                   if (id[u]!=edge[k].id)                                   {                                           v=edge[k].v;                                           if (dfn[v]>oo)                                           {                                                   dfn[v]=dfn[u]+1;                                                 id[v]=edge[k].id;                                                 Q.push(v);                                           }else                                                 ans=min(ans,dfn[v]+dfn[u]+1);                                   }                        }                 }                printf("Case %d: ",cases);                if (ans==oo) puts("impossible");                        else printf("%d\n",ans);       }       return 0;}