树（最小乘积生成树,克鲁斯卡尔算法）：BOI timeismoney

The NetLine company wants to offer broadband internet to N towns. For this, it suffices to construct
a network of N-1 broadband links between the towns, with the property that a message can travel
from any town to any other town on this network. NetLine has already identified all pairs of towns
between which a direct link can be constructed. For each such possible link, they know the cost and
the time it would take to construct the link.
The company is interested in minimizing both the total amount of time (links are built one at a time)
and the total amount of money spent to build the entire network. Since they couldn’t decide among
the two criteria, they decided to use the following formula to evaluate the value of a network:
SumTime = sum of times spent to construct the chosen links
SumMoney = sum of the money spent to construct the chosen links
V = SumTime * SumMoney
Task
Find a list of N-1 links to build, which minimizes the value V.
Description of input
The first line of input contains integers N – the number of towns and M – the number of pairs of
towns which can be connected. The towns are numbered starting from 0 to N-1. Each of the next M
lines contain four integers x, y, t and c – meaning town x can be connected to town y in time t and
with cost c.
Description of output
In the first line of output print two numbers: the total time (SumTime) and total money (Sum-
Money) used in the optimal solution (the one with minimal value V), separated by one space. The
next N-1 lines describe the links to be constructed. Each line contains a pair of numbers (x,y) describing
a link to be build (which must be among the possible links described in the input file). The
pairs can be printed out in any order. When multiple solutions exist, you may print any of them.

Constraints

· 1 ≤ N ≤ 200
· 1 ≤ M ≤ 10 000
· 0 ≤ x,y ≤ N-1
· 1 ≤ t,c ≤ 255
· One test has M = N - 1
· 40% of the tests will have for each possible link t = c
Example
timeismoney.in
5 7
0 1 161 79
0 2 161 15
0 3 13 153
1 4 142 183
2 4 236 80
3 4 40 241
2 1 65 92

timeismoney.out

279 501
2 1
0 3
0 2
3 4

　　方案啥的很好解决，就不写了。

　　和HNOI2014画框有类似的思想。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=210;
const int maxm=10010;
int fa[maxn],a[maxm],b[maxm];
int u[maxm],v[maxm],n,m;

struct Node{
    int x,y,z,id;
    Node(int a=0,int b=0,int c=0,int d=0){
        x=a;y=b;z=c;id=d;
    }
    friend bool operator <(Node a,Node b){
        return a.z<b.z;
    }
}p[maxm];

struct Point{
    int x,y;
    Point(int a=0,int b=0){
        x=a;y=b;
    }
    friend bool operator ==(Point a,Point b){
        return a.x==b.x&&a.y==b.y;
    }
}lo,hi;

int Find(int x){
    return fa[x]==x?x:fa[x]=Find(fa[x]);
}

Point Get_Ans(){
    sort(p+1,p+m+1);Point ret(0,0);
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=1;i<=m;i++){
        int x=p[i].x,y=p[i].y;
        if(Find(x)!=Find(y)){
            ret.x+=a[p[i].id];
            ret.y+=b[p[i].id];
            fa[Find(y)]=Find(x);
        }
    }
    return ret;
}

Point Solve(Point l,Point r){
    for(int i=1;i<=m;i++)
        p[i]=Node(u[i],v[i],b[i]*(r.x-l.x)-a[i]*(r.y-l.y),i);
    Point mid=Get_Ans();
    if(mid==l||mid==r)return l.x*l.y<r.x*r.y?l:r;
    l=Solve(l,mid);r=Solve(mid,r);
    return l.x*l.y<r.x*r.y?l:r;    
}

int main(){
    freopen("timeismoney.in","r",stdin);
    freopen("timeismoney.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u[i],&v[i]);u[i]+=1;v[i]+=1;
        scanf("%d%d",&a[i],&b[i]);
    }
    
    for(int i=1;i<=m;i++)p[i]=Node(u[i],v[i],a[i],i);lo=Get_Ans();
    for(int i=1;i<=m;i++)p[i]=Node(u[i],v[i],b[i],i);hi=Get_Ans();    
    Point ans=Solve(lo,hi);
    printf("%d %d
",ans.x,ans.y);    
    return 0;
}