cf [disjoint-set]

http://codeforces.com/contest/1245/problem/D

Meaning of the questions is: you need to let all cities have electricity, you also see the city build power plants so that he has power, and you can link him with other cities, making it electrically

solve:

We can now build their own power plants each city as well as its own costs and other cities keep up with the structure, row a sequence, and then check even set up.

#include<bits/stdc++.h>
#define name ch-48
#define pd putchar(' ')
#define pn putchar('\n')
#define pb push_back
#define debug(args...) cout<<#args<<"->"<<args<<endl
#define bug cout<<"************"
using namespace std;
template <typename T>
void read(T &res) {
    bool flag=false;char ch;
    while(!isdigit(ch=getchar())) (ch=='-')&&(flag=true);
    for (res = name; isdigit (ch = getchar ()); res = (res << 1) + (res << 3) + name);
    flag&&(res=-res);
}
template <typename T>
void write(T x) {
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
typedef long long ll;
typedef long double ld;
const int maxn=2000+10;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const double alpha=0.7;
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
struct node {
    you and, v;
    ll w;
    node(){}
    node(int u,int v,ll w):u(u),v(v),w(w){}
    bool operator<(const node&a) {
        return w<a.w;
    }
}e[maxn*maxn+maxn];
struct a {
    ll x,y;
    int pos;
}a[maxn];
ll c[maxn],k[maxn];
int f[maxn];
vector<int >vec1;
vector<pii >vec2;
int getf(int v) {
    return f [v] == v v: f [v] = getf (f [v]);?
}
int main ()
{
    int n;
    read(n);
    for(int i=1;i<=n;i++)
        f[i]=i;
    for(int i=1;i<=n;i++) {
        read(a[i].x),read(a[i].y);
//        a[i].pos=i;
    }
    for(int i=1;i<=n;i++)
        read(c[i]);
    for(int i=1;i<=n;i++)
        read(k[i]);
    int cnt=0;
    for(int i=1;i<=n;i++){
        e[++cnt]=node(0,i,c[i]);
        for(int j=i+1;j<=n;j++)
            e[++cnt]=node(i,j,(k[i]+k[j])*(abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)));
    }
    sort (e + 1 and + 1 + cnt);
    int num = 0;
    ll sum=0;
    for(int i=1;i<=cnt;i++){
        int u=e[i].u,v=e[i].v;
        int a=getf(u);
        int b=getf(v);
        if (a! = b) {// only need to calculate a different father, for example, three points you sure three sides of the triangular frame, but in fact only on the line where the two sides, the third certainly be recognized We have a common father
            f[b]=a;
            if(u==0)
                vec1.pb(v);
            else {
                vec2.pb(mp(u,v));
            }
            num ++; // Whether you build your own base station or connection, because the connection is the default front that already have electricity, only a plus
            sum+=e[i].w;
            if (num == n) break; // to ensure that the n cities are bright
        }
    }
    write(sum);pn;
    write(vec1.size());pn;
    if(vec1.size()) {
        for(int i=0;i<vec1.size();i++)
            write(vec1[i]),pd;
        pn;
    }
    write(vec2.size());pn;
    if(vec2.size()) {
        for(int i=0;i<vec2.size();i++)
            write(vec2[i].fi),pd,write(vec2[i].se),pn;;
        pn;
    }
    return 0;
}