6-10. 关键活动

所谓关键路径就是DAG的最长路径，用dfs找

#include<bitset>
#include<iostream>
#include<set>
#include<algorithm>
using namespace std;

const int N=103,INF=1<<30,src=0,dst=N-1;
int n,m,pre[N],dis[N][N],nth[N][N],maxlen,nocycle=true;
struct road{
  int src,dst,nth;
  road(int a,int b,int c):src(a),dst(b),nth(c){}
  bool operator<(const road&x)const{
    return src!=x.src?src<x.src:nth>x.nth;}
};
set<road>path;
void storepath(int k){
  if(pre[k]!=src)storepath(pre[k]);
  if(pre[k]!=src&&k!=dst)
    path.insert(road(pre[k],k,nth[pre[k]][k]));
}
void dfs(int k,int len){
  static bitset<N>used;
  used.set(k);
  if(k==dst){
    if(maxlen<len){
      maxlen=len;
      path.clear();
      storepath(k);
    }else if(maxlen==len)
      storepath(k);
  }else{
    for(int i=0;i<N;++i)
      if(dis[k][i]!=INF){
	if(used[i]) nocycle=false;
	pre[i]=k;
	dfs(i,len+dis[k][i]);}
  }
  used.reset(k);
}
void initial(){
  fill_n(*dis,N*N,INF);
  fill_n(pre,N,-1);
  vector<bool>nopre(N,true),nonext(N,true);
  cin>>n>>m;
  for(int i=1;i<=m;++i){
    int a,b,c;cin>>a>>b>>c;
    dis[a][b]=c;
    nth[a][b]=i;
    nopre[b]=false,nonext[a]=false;
  }
  for(int i=1;i<=n;++i){
    if(nopre[i])dis[src][i]=0;
    if(nonext[i])dis[i][dst]=0;
  }
}
int main(){
  initial();
  dfs(src,0);
  if(nocycle){
    cout<<maxlen<<endl;
    for(auto it=path.begin();it!=path.end();++it)
      cout<<it->src<<"->"<<it->dst<<endl;
  }else cout<<0;
  return 0;
}