description
某国有n座城市,这些城市之间通过m条单向道路相连,已知每条道路的长度。
不过,小X只对其中k座城市感兴趣。
为了更好地规划模拟旅行路线,提升模拟旅行的体验,小X想要知道他感兴趣的城市之间两两最短路的最小值(即在他感兴趣的城市中,最近的一对的最短距离)。
作为一个肥宅,小X根本懒得写程序来解决这道问题,于是他把这个问题丢给了你。
J 国有 nn 座城市,这些城市之间通过 mm 条单向道路相连,已知每条道路的长度。
一次,居住在 J 国的 Rainbow 邀请 Vani 来作客。不过,作为一名资深的旅行者,Vani 只对 J 国的 kk 座历史悠久、自然风景独特的城市感兴趣。
为了提升旅行的体验,Vani 想要知道他感兴趣的城市之间「两两最短路」的最小值(即在他感兴趣的城市中,最近的一对的最短距离)。
也许下面的剧情你已经猜到了——Vani 这几天还要忙着去其他地方游山玩水,就请你帮他解决这个问题吧。
analysis
-
对于所有关键点开始跑最短路,原图和反图分别跑一次,记为(dis_i,dis1_i)
-
记录下(dij)里每个点从哪一个关键点转移得来,记为(from_i,from1_i)
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枚举一条边,如果边的两端点的(from_x≠from_y),此时可以拿(dis_x+dis1_y+len)更新答案
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因为(from_x≠from_y),所以两端点是从不同关键点走到,说明该边可能在两个关键点的最短路上
-
还有一种更妙的做法,不过(SPFA)弄不了,我还没实现
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建立超级源向所有关键点连(0)边,然后从超级源跑最短路和次短路
-
关键点的最短路当然是(0),次短路的意义是什么呢
-
其实某关键点的次短路一定是经过另一个关键点走到它的最短路
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答案取次短路的最小值就可以了,感觉确实妙
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXN 300005
#define MAXM 3000005
#define ha 1926081719491001
#define ll long long
#define reg register ll
#define max(x,y) ((x>y)?(x):(y))
#define min(x,y) ((x<y)?(x):(y))
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=last[a];i;i=next[i])
#define rep1(i,a) for (reg i=last1[a];i;i=next1[i])
using namespace std;
ll last[MAXM],next[MAXM],tov[MAXM],len[MAXM];
ll last1[MAXM],next1[MAXM],tov1[MAXM],len1[MAXM];
ll a[MAXN],dis[MAXN],dis1[MAXN],from[MAXN],from1[MAXN];
ll n,m,k,tot,tot1,ans=ha;
ll edge[MAXM][3];
bool bz[MAXN];
struct node
{
ll x,y;
bool operator<(const node &a)const{return a.y<y;}
};
priority_queue<node>q;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline void link(ll x,ll y,ll z){next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z;}
inline void link1(ll x,ll y,ll z){next1[++tot1]=last1[x],last1[x]=tot1,tov1[tot1]=y,len1[tot1]=z;}
inline void dijkstra()
{
memset(dis,100,sizeof(dis));
memset(bz,0,sizeof(bz));
fo(i,1,k)q.push((node){a[i],dis[a[i]]=0}),from[a[i]]=a[i];
while (!q.empty())
{
node now=q.top();q.pop();
if (bz[now.x])continue;bz[now.x]=1;
rep(i,now.x)if (dis[now.x]+len[i]<dis[tov[i]])
{
dis[tov[i]]=dis[now.x]+len[i],from[tov[i]]=from[now.x];
q.push((node){tov[i],dis[tov[i]]});
}
}
}
inline void dijkstra1()
{
memset(dis1,100,sizeof(dis1));
memset(bz,0,sizeof(bz));
fo(i,1,k)q.push((node){a[i],dis1[a[i]]=0}),from1[a[i]]=a[i];
while (!q.empty())
{
node now=q.top();q.pop();
if (bz[now.x])continue;bz[now.x]=1;
rep1(i,now.x)if (dis1[now.x]+len1[i]<dis1[tov1[i]])
{
dis1[tov1[i]]=dis1[now.x]+len1[i],from1[tov1[i]]=from1[now.x];
q.push((node){tov1[i],dis1[tov1[i]]});
}
}
}
int main()
{
freopen("tour.in","r",stdin);
freopen("tour.out","w",stdout);
n=read(),m=read(),k=read();
fo(i,1,m)
{
ll x=read(),y=read(),z=read();
edge[i][0]=x,edge[i][1]=y,edge[i][2]=z;
link(x,y,z),link1(y,x,z);
}
fo(i,1,k)a[i]=read();
dijkstra(),dijkstra1();
fo(i,1,m)if (from[edge[i][0]]!=from1[edge[i][1]])
ans=min(ans,dis[edge[i][0]]+dis1[edge[i][1]]+edge[i][2]);
printf("%lld
",ans);
return 0;
}
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXN 300005
#define MAXM 3000005
#define ha 1926081719491001
#define ll long long
#define reg register ll
#define max(x,y) ((x>y)?(x):(y))
#define min(x,y) ((x<y)?(x):(y))
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=last[a];i;i=next[i])
#define rep1(i,a) for (reg i=last1[a];i;i=next1[i])
using namespace std;
ll last[MAXM],next[MAXM],tov[MAXM],len[MAXM];
ll last1[MAXM],next1[MAXM],tov1[MAXM],len1[MAXM];
ll a[MAXN],dis[MAXN],dis1[MAXN],from[MAXN],from1[MAXN];
ll n,m,k,T,tot,tot1,ans=ha;
ll edge[MAXM][3];
bool bz[MAXN];
struct node
{
ll x,y;
bool operator<(const node &a)const{return a.y<y;}
};
priority_queue<node>q;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline void link(ll x,ll y,ll z){next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z;}
inline void link1(ll x,ll y,ll z){next1[++tot1]=last1[x],last1[x]=tot1,tov1[tot1]=y,len1[tot1]=z;}
inline void dijkstra()
{
memset(dis,100,sizeof(dis));
memset(bz,0,sizeof(bz));
fo(i,1,k)q.push((node){a[i],dis[a[i]]=0}),from[a[i]]=a[i];
while (!q.empty())
{
node now=q.top();q.pop();
if (bz[now.x])continue;bz[now.x]=1;
rep(i,now.x)if (dis[now.x]+len[i]<dis[tov[i]])
dis[tov[i]]=dis[now.x]+len[i],from[tov[i]]=from[now.x],q.push((node){tov[i],dis[tov[i]]});
}
}
inline void dijkstra1()
{
memset(dis1,100,sizeof(dis1));
memset(bz,0,sizeof(bz));
fo(i,1,k)q.push((node){a[i],dis1[a[i]]=0}),from1[a[i]]=a[i];
while (!q.empty())
{
node now=q.top();q.pop();
if (bz[now.x])continue;bz[now.x]=1;
rep1(i,now.x)if (dis1[now.x]+len1[i]<dis1[tov1[i]])
dis1[tov1[i]]=dis1[now.x]+len1[i],from1[tov1[i]]=from1[now.x],q.push((node){tov1[i],dis1[tov1[i]]});
}
}
int main()
{
T=read();
while (T--)
{
memset(last,0,sizeof(last)),memset(last1,0,sizeof(last1));
memset(next,0,sizeof(next)),memset(next1,0,sizeof(next1));
memset(tov,0,sizeof(tov)),memset(tov1,0,sizeof(tov1)),tot=0;
memset(len,0,sizeof(len)),memset(len1,0,sizeof(len1)),tot1=0;
memset(from,0,sizeof(from)),memset(from1,0,sizeof(from1));
n=read(),m=read(),k=read(),ans=ha;
fo(i,1,m)
{
ll x=read(),y=read(),z=read();
edge[i][0]=x,edge[i][1]=y,edge[i][2]=z;
link(x,y,z),link1(y,x,z);
}
fo(i,1,k)a[i]=read();
dijkstra(),dijkstra1();
fo(i,1,m)if (from[edge[i][0]]!=from1[edge[i][1]])
ans=min(ans,dis[edge[i][0]]+dis1[edge[i][1]]+edge[i][2]);
printf("%lld
",ans);
}
return 0;
}