題解

最小乘積生成樹！

我們把，x的總和和y的總和作為x坐標和y左邊，畫在坐標系上

我們選擇兩個初始點，一個是最靠近y軸的A，也就是x總和最小，一個是最靠近x軸的B，也就是y總和最小
連接兩條直線，在這條直線上面的點都不用考慮了

我們選一個離直線最遠的點C，且在直線下方，我們用叉積考慮這個東西，也就是……面積最大！我們如果用最小生成樹的話，只要讓面積是負的就好了
推一下式子，發現是\((A.y - B.y) * C.x + (B.x - A.x) * C.y\)我們發現就是把邊設置成
\((A.y - B.y) * E[i].c + (B.x - A.x) * E[i].t\)做一遍最小生成樹

找到C點後遞歸處理A,C和C,B即可

邊界是兩點連線下方沒有點也就是叉積大於等於0

代碼

#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(‘\n‘)
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\n",ans.x,ans.y);
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

【BZOJ】2395: [Balkan 2011]Timeismoney