洛谷 1821 [USACO07FEB]银牛派对Silver Cow Party

【题解】

　　其实解法

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define LL long long
 #define rg register
 #define N 200010
 using namespace std;
 int n,m,s,ans,tot,last[N],dis[N],dis2[N],pos[N];
 struct rec{
     int u,v,d;
 }r[N<<];
 struct edge{
     int to,pre,dis;
 }e[N<<];
 struct heap{
     int poi,dis;
 }h[N];
 inline int read(){
     int k=,f=; char c=getchar();
     while(c<''||c>'')c=='-'&&(f=-),c=getchar();
     while(''<=c&&c<='')k=k*+c-'',c=getchar();
     return k*f;
 }
 inline void up(int x){
     int fa;
     while((fa=(x>>))&&h[fa].dis>h[x].dis){
         swap(h[fa],h[x]); swap(pos[h[fa].poi],pos[h[x].poi]);
         x=fa;
     }
 }
 inline void down(int x){
     int son;
     while((son=(x<<))<=tot){
         if(son<tot&&h[son].dis>h[son+].dis) son++;
         if(h[son].dis<h[x].dis){
             swap(h[son],h[x]); swap(pos[h[son].poi],pos[h[x].poi]);
             x=son;
         }
         else return;
     }
 }
 inline void dijkstra(int x){
     for(rg int i=;i<=n*;i++) dis[i]=1e9;
     h[tot=pos[x]=]=(heap){x,dis[x]=};
     while(tot){
         int now=h[].poi; h[]=h[tot--]; if(tot) down();
         for(rg int i=last[now],to;i;i=e[i].pre)
         if(dis[to=e[i].to]>dis[now]+e[i].dis){
             dis[to]=dis[now]+e[i].dis;
             if(!pos[to]) h[pos[to]=++tot]=(heap){to,dis[to]};
             else h[pos[to]].dis=dis[to];
             up(pos[to]);
         }
         pos[now]=;
     }
 }
 int main(){
     n=read(); m=read(); s=read();
     for(rg int i=;i<=m;i++){
         int u=read(),v=read(),d=read();
         r[i]=(rec){u,v,d};
         e[++tot]=(edge){v,last[u],d}; last[u]=tot;
     }
     dijkstra(s);
     for(rg int i=;i<=n;i++) dis2[i]=dis[i];
     memset(last,,sizeof(last));
     for(rg int i=;i<=m;i++){
         int u=r[i].u,v=r[i].v,d=r[i].d;
         e[++tot]=(edge){u,last[v],d}; last[v]=tot;
     }
     dijkstra(s);
     for(rg int i=;i<=n;i++) ans=max(ans,dis[i]+dis2[i]);
     printf("%d\n",ans);
     return ;
 }

就是先做一遍最短路，把所有边反向，再做一遍最短路，最后找两次的dis之和的最大值。