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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();
>
> }
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