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最长路是NP的。。。In Reply To:求负权的单源最短路是不是等于求正权的单源最长路?我的代码WA了-_-|| Posted by:animate at 2005-10-19 15:19:12 > #include<cstdio> > #include<iostream> > using namespace std; > > int n,m,s; > int E[100200][3]; > int D[50100]; > > void BellmanFord(){ > int i,j,k; > for(i=s;i<=m;i++)D[i]=1; > D[s]=0; > for(k=1;k<=m;k++){ > > for(j=0;j<n;j++){ > int u=E[j][0],v=E[j][1],e=E[j][2]; > if(D[u]!=1&&D[v]>D[u]+e){ > D[v]=D[u]+e; > } > } > > } > > printf("%d\n",-D[m]); > } > > int main(){ > //freopen("in.txt","r",stdin); > scanf("%d",&n); > int i; > int u,v,e; > m=-1;s=-1; > > for(i=0;i<n;i++){ > scanf("%d%d%d",&u,&v,&e); > E[i][0]=u; > if(s==-1||s>u)s=u; > E[i][1]=v+1; > if(v+1>m)m=v+1; > E[i][2]=-e; > } > > for(i=s;i<m;i++){ > E[n][0]=i; > E[n][1]=i+1; > E[n][2]=0; > n++; > } > > BellmanFord(); > > } Followed by: Post your reply here: |
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