[Codeforces 1245D] Shichikuji and Power Grid (minimum spanning tree)

[Codeforces 1245D] Shichikuji and Power Grid (minimum spanning tree)

Face questions

There are n cities, coordinates \ ((x_i, y_i) \) , there are two coefficients \ (C_i, K_i \) . Need to build power stations cost in each city \ (C_i \) . If you do not build power stations , let the city electricity, we need to have communication with the city power plant. the cost of even a undirected edge between i and j is \ ((K_i + k_j) \) * Manhattan distance between the two cities. seek to make the minimum cost of each city are energized, and the output of any one program.

analysis

The cost of each point in turn the election into a virtual point of origin to the edge, this routine is very common, but it is the first time I saw in the minimum spanning tree problem.

Two of the city between the two linkage side, the right side is calculated by reference to the subject in. Then create a virtual source point to each city \ (i \) even side, the right side is \ (C_i \) . Figure for the entire run a minimum spanning tree can be.

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define maxn 2000
using namespace std;
typedef long long ll;
int n; 
struct node{
    int x;
    int y;
    int c;
    int k;
}a[maxn+5];
inline ll dist(ll x1,ll y1,ll x2,ll y2){
    return abs(x1-x2)+abs(y1-y2);
} 

struct edge{
    int from;
    int to;
    ll len;
    int next;
}E[maxn*maxn+5];
int head[maxn+5];
int sz=0;
void add_edge(int u,int v,ll w){
    sz++;
    E[sz].from=u;
    E[sz].to=v;
    E[sz].len=w;
    E[sz].next=head[u];
    head[u]=sz;
} 
bool cmp(edge p,edge q){
    return p.len<q.len;
}

int fa[maxn+5];
int find(int x){
    if(fa[x]==x) return x;
    else return fa[x]=find(fa[x]);
}

vector< pair<int,int> >res;
ll kruskal(){
    ll ans=0;
    for(int i=1;i<=n+1;i++) fa[i]=i;
    sort(E+1,E+1+sz,cmp);
    for(int i=1;i<=sz;i++){
        int u=E[i].from,v=E[i].to;
        int fu=find(u),fv=find(v);
        if(fu!=fv){
            ans+=E[i].len;
            fa[fu]=fv;
            res.push_back(make_pair(E[i].from,E[i].to));
        } 
    }
    return ans;
}

vector<int>poi;//poi~ 
vector< pair<int,int> >edg;
int main(){
    int st;
    scanf("%d",&n);
    st=n+1;
    for(int i=1;i<=n;i++) scanf("%d %d",&a[i].x,&a[i].y);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i].c);
        add_edge(st,i,a[i].c);
    }
    for(int i=1;i<=n;i++) scanf("%d",&a[i].k);
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            add_edge(i,j,(a[i].k+a[j].k)*dist(a[i].x,a[i].y,a[j].x,a[j].y));
        }
    }
    printf("%I64d\n",kruskal());
    for(auto ed : res){
        if(ed.first==st) poi.push_back(ed.second);
        else edg.push_back(ed);
    }
    printf("%d\n",poi.size());
    for(int x : poi) printf("%d ",x);
    printf("\n");
    printf("%d\n",edg.size());
    for(auto x : edg) printf("%d %d\n",x.first,x.second); 
}