Problem E. Opening Portals
首先根据传送门的性质，如果所有点都是传送门的话那么结果就是该图的最小生成树。
对于只有其中 k 个结点是传送门的图，只要在原算法的基础上稍作修改即可。
具体，对每个点求出 P[i] 和 D[i] 。（表示距离这个点最近的传送门和其距离。。。
之后对每条边，再根据 D[x] + D[y] + w 作为关键字跑最小生成树。。
以上分别用一次多源 spfa()，和稍作修改的 Kruskal() 即可。
E. Opening Portals
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Pavel plays a famous computer game. A player is responsible for a whole country and he can travel there freely, complete quests and earn experience.
This country has n cities connected by m bidirectional
roads of different lengths so that it is possible to get from any city to any other one. There are portals in k of these cities. At the beginning of the
game all portals are closed. When a player visits a portal city, the portal opens. Strange as it is, one can teleport from an open portal to an open one. The teleportation takes no time and that enables the player to travel quickly between rather remote regions
of the country.
At the beginning of the game Pavel is in city number 1. He wants to open all portals as quickly as possible. How much time will he need for that?
Input
The first line contains two space-separated integers n and m (1?≤?n?≤?105, 0?≤?m?≤?105)
that show how many cities and roads are in the game.
Each of the next m lines contains the description of a road as three space-separated integers xi, yi, wi (1?≤?xi,?yi?≤?n, xi?≠?yi, 1?≤?wi?≤?109)
— the numbers of the cities connected by the i-th road and the time needed to go from one city to the other one by this road. Any two cities are connected
by no more than one road. It is guaranteed that we can get from any city to any other one, moving along the roads of the country.
The next line contains integer k (1?≤?k?≤?n)
— the number of portals.
The next line contains k space-separated integers p1, p2,
..., pk — numbers
of the cities with installed portals. Each city has no more than one portal.
Output
Print a single number — the minimum time a player needs to open all portals.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams
or the %I64dspecifier.
Sample test(s)
input
3 3 1 2 1 1 3 1 2 3 1 3 1 2 3
output
2
input
4 3 1 2 1 2 3 5 2 4 10 3 2 3 4
output
16
input
4 3 1 2 1000000000 2 3 1000000000 3 4 1000000000 4 1 2 3 4
output
3000000000
Note
In the second sample the player has to come to city 2, open a portal there, then go to city 3,
open a portal there, teleport back to city 2and finally finish the journey in city 4.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=100100;
typedef long long int LL;
int n,m,c;
struct Edge
{
int from,to,next;
LL weight;
}edge[3*maxn];
int Adj[maxn],Size;
int p[maxn],cq[maxn];
LL dist[maxn];
bool inq[maxn];
void init()
{
memset(Adj,-1,sizeof(Adj)); Size=0;
}
void Add_Edge(int u,int v,LL w)
{
edge[Size].from=u;
edge[Size].to=v;
edge[Size].next=Adj[u];
edge[Size].weight=w;
Adj[u]=Size++;
}
bool spfa()
{
memset(dist,63,sizeof(dist));
memset(p,0,sizeof(p));
memset(cq,0,sizeof(cq));
memset(inq,false,sizeof(inq));
int k=c;
queue<int> q;
for(int i=0;i<k;i++)
{
int x;
scanf("%d",&x);
dist[x]=0; p[x]=x;
q.push(x); cq[x]=1; inq[x]=true;
}
while(!q.empty())
{
int u=q.front(); q.pop();
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(dist[v]>dist[u]+edge[i].weight)
{
dist[v]=dist[u]+edge[i].weight;
p[v]=p[u];
if(!inq[v])
{
inq[v]=true;
cq[v]++;
if(cq[v]>=n) return false;
q.push(v);
}
}
}
inq[u]=false;
}
return true;
}
struct BA
{
int x,y;
LL w;
}bian[3*maxn];
bool cmp(BA a,BA b)
{
return a.w<b.w;
}
int fa[maxn];
int find(int x)
{
if(fa[x]==x) return x;
return fa[x]=find(fa[x]);
}
int main()
{
scanf("%d%d",&n,&m);
init();
for(int i=0;i<m;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
Add_Edge(a,b,c);
Add_Edge(b,a,c);
}
scanf("%d",&c);
spfa();
for(int i=0;i<=n+10;i++) fa[i]=i;
for(int i=0;i<Size;i++)
{
int u=edge[i].from,v=edge[i].to;
bian[i].x=p[u],bian[i].y=p[v];
bian[i].w=dist[u]+dist[v]+edge[i].weight;
}
sort(bian,bian+Size,cmp);
int cnt=1;
LL ans=dist[1];
for(int i=0;i<Size&&cnt<c;i++)
{
int u=bian[i].x,v=bian[i].y;
LL w=bian[i].w;
if(u==v) continue;
int U=find(u),V=find(v);
if(U==V) continue;
else
{
ans+=w;
fa[U]=V;
cnt++;
}
if(cnt>=c) break;
}
cout<<ans<<endl;
return 0;
}
Codeforces 196 E. Opening Portals