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<<1];
struct edge{
    int to,pre,dis;
}e[N<<1];
struct heap{
    int poi,dis;
}h[N];
inline int read(){
    int k=0,f=1; char c=getchar();
    while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
    while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar();
    return k*f;
}
inline void up(int x){
    int fa;
    while((fa=(x>>1))&&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<<1))<=tot){
        if(son<tot&&h[son].dis>h[son+1].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=1;i<=n*2;i++) dis[i]=1e9;
    h[tot=pos[x]=1]=(heap){x,dis[x]=0};
    while(tot){
        int now=h[1].poi; h[1]=h[tot--]; if(tot) down(1);
        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]=0;
    }
}
int main(){
    n=read(); m=read(); s=read();
    for(rg int i=1;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=1;i<=n;i++) dis2[i]=dis[i];
    memset(last,0,sizeof(last));
    for(rg int i=1;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=1;i<=n;i++) ans=max(ans,dis[i]+dis2[i]);
    printf("%d
",ans);
    return 0;
}

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。