【BZOJ】2395: [Balkan 2011]Timeismoney

题解

最小乘积生成树！

我们把，x的总和和y的总和作为x坐标和y左边，画在坐标系上

我们选择两个初始点，一个是最靠近y轴的A，也就是x总和最小，一个是最靠近x轴的B，也就是y总和最小
连接两条直线，在这条直线上面的点都不用考虑了

我们选一个离直线最远的点C，且在直线下方，我们用叉积考虑这个东西，也就是……面积最大！我们如果用最小生成树的话，只要让面积是负的就好了
推一下式子，发现是((A.y - B.y) * C.x + (B.x - A.x) * C.y)我们发现就是把边设置成
((A.y - B.y) * E[i].c + (B.x - A.x) * E[i].t)做一遍最小生成树

找到C点后递归处理A,C和C,B即可

边界是两点连线下方没有点也就是叉积大于等于0

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <vector>
#include <set>
//#define ivorysi
#define eps 1e-8
#define mo 974711
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 10005
#define space putchar(' ')
#define enter putchar('
')
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef unsigned long long u64;
typedef double db;
const int64 MOD = 1000000007;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) putchar('-');
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
struct Point {
    int64 x,y;
    int64 v;
    Point(){};
    Point(int64 _x,int64 _y) {
	x = _x;y = _y;v = x * y;
    }
    friend bool operator < (const Point &a,const Point &b) {
	return a.v < b.v || (a.v == b.v && a.x < b.x);
    }
}ans;
struct Edge {
    int u,v;
    int64 c,t,w;
    Edge(){}
    Edge(int _u,int _v,int64 _c,int64 _t) {
	u = _u;v = _v;c = _c;t = _t;
    }
    friend bool operator < (const Edge &a,const Edge &b) {
	return a.w < b.w || (a.w == b.w && a.c < b.c);
    }
}E[MAXN];
int fa[205];
int getfa(int u) {
    return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
Point kruskal() {
    sort(E + 1,E + M + 1);
    Point res = Point(0,0);
    for(int i = 1 ; i <= N ; ++i) fa[i] = i;
    for(int i = 1 ; i <= M ; ++i) {
	if(getfa(E[i].u) != getfa(E[i].v)) {
	    fa[getfa(E[i].u)] = getfa(E[i].v);
	    res.x += E[i].c;res.y += E[i].t;
	}
    }
    res.v = res.x * res.y;
    if(res < ans) ans = res;
    return res;
}
void Work(Point A,Point B) {
    for(int i = 1 ; i <= M ; ++i) {
	E[i].w = (A.y - B.y) * E[i].c + (B.x - A.x) * E[i].t;
    }
    Point r = kruskal();
    if((A.x - r.x) * (B.y - r.y) - (A.y - r.y) * (B.x - r.x) >= 0) return;
    Work(A,r);
    Work(r,B);
}
void Solve() {
    read(N);read(M);
    int u,v;
    int64 c,t;
    for(int i = 1 ; i <= M ; ++i) {
	read(u);read(v);read(c);read(t);
	++u;++v;
	E[i] = Edge(u,v,c,t);
    }
    ans.v = 1e18;
    for(int i = 1 ; i <= M ; ++i) {
	E[i].w = E[i].c;
    }
    Point A = kruskal();
    for(int i = 1 ; i <= M ; ++i) {
	E[i].w = E[i].t;
    }
    Point B = kruskal();
    Work(A,B);
    printf("%lld %lld
",ans.x,ans.y);
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}