BZOJ 2395 Time is Money

AC通道：http://www.lydsy.com/JudgeOnline/problem.php?id=2395

 

直接上题解：http://www.cnblogs.com/autsky-jadek/p/3959446.html

很好看懂，感受到了画图的力量！点对投射到二维空间中去，感觉就会明了很多的样子...

不过复杂度什么的...虽然不能直接算出来，不过跑得很快。

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=210;
const int maxm=10010;

struct Edge{
    int u,v,w,t,c;
}e[maxm];

struct Point{
    int x,y;
    Point(){x=0,y=0;}
    Point(int a,int b){x=a,y=b;}
}ans;

typedef Point Vector;
typedef long long ll;

Vector operator - (const Point &A,const Point &B){
    return Vector(A.x-B.x,A.y-B.y);
}

ll Cross(const Vector &A,const Vector &B){
    return (ll)A.x*B.y-(ll)A.y*B.x;
}

bool cmp(const Edge &A,const Edge &B){
    return A.w<B.w;
}

int n,m;
int p[maxn];

int Find(int x){
    int r=x,pre;
    while(p[r]!=r) r=p[r];
    while(x!=r)
        pre=p[x],p[x]=r,x=pre;
    return x;
}

Point Kruscal(){
    Point res;
    int cot=1,fx,fy;
    for(int i=1;i<=n;i++) p[i]=i; 
    for(int i=1;i<=m && cot<n;i++){
        fx=Find(e[i].u),fy=Find(e[i].v);
        if(fx!=fy)
            cot++,res.x+=e[i].c,res.y+=e[i].t,p[fx]=fy;
    }
    if((ll)ans.x*ans.y>(ll)res.x*res.y || ((ll)ans.x*ans.y==(ll)res.x*res.y && ans.x>res.x)) ans=res;
    return res;
}

void Div(Point A,Point B){
    for(int i=1;i<=m;i++)
        e[i].w=e[i].c*(A.y-B.y)+e[i].t*(B.x-A.x);
    sort(e+1,e+m+1,cmp);
    Point C=Kruscal();
    if(Cross(B-A,C-A)>=0) return ;
    Div(A,C);Div(C,B);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("2395.in","r",stdin);
    freopen("2395.out","w",stdout);
#endif
    
    Point A,B;
    ans.x=1e9,ans.y=1e9;

    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].c,&e[i].t);
        e[i].u++,e[i].v++;
    }
    for(int i=1;i<=m;i++) e[i].w=e[i].c;
    sort(e+1,e+m+1,cmp);
    A=Kruscal();
    for(int i=1;i<=m;i++) e[i].w=e[i].t;
    sort(e+1,e+m+1,cmp);
    B=Kruscal();
    Div(A,B);
    
    printf("%d %d",ans.x,ans.y);
    
    return 0;
}