zoukankan      html  css  js  c++  java
  • bzoj4578: [Usaco2016 OPen]Splitting the Field

    bzoj4578: [Usaco2016 OPen]Splitting the Field

    Time Limit: 10 Sec  Memory Limit: 128 MB
    Submit: 56  Solved: 15
    [Submit][Status][Discuss]

    Description

    Farmer John's N cows (3≤N≤50,000) are all located at distinct positions in his two-dimensional fie
    ld. FJ wants to enclose all of the cows with a rectangular fence whose sides are parallel to the x a
    nd y axes, and he wants this fence to be as small as possible so that it contains everycow (cows on 
    the boundary are allowed).FJ is unfortunately on a tight budget due to low milk production last quar
    ter. He would therefore like to enclose a smaller area to reduce maintenance costs, and the only way
     he can see to do this is by building two enclosures instead of one. Please help him compute how muc
    h less area he needs to enclose, in total, by using two enclosures instead of one. Like the original
     enclosure, the two enclosures must collectively contain all the cows (with cows on boundaries allow
    ed), and they must have sides parallel to the x and y axes. The two enclosures are notallowed to ove
    rlap -- not even on their boundaries. Note that enclosures of zero area are legal, for example if an
     enclosure has zero width and/or zero height.

    Input

    The first line of input contains N. The nextNN lines each contain two integers specifying the location of a cow. Cow locations are positive integers in the range 1…1,000,000,000

    Output

    Write a single integer specifying amount of total area FJ can save by using two enclosures instead o
    f one.

    Sample Input

    6
    4 2
    8 10
    1 1
    9 12
    14 7
    2 3

    Sample Output

    107

    HINT

     

    Source

    题意:给出若干个点 问如果用两个矩形覆盖所有点会比用一个矩形覆盖所有点节省多少面积

    题解:。。。好坑啊。。。矩形在边界上似乎是可以相交的然而题目说不行。。。然后面积应该是矩形的 (x2-x1) * (y2-y1) ..是不用加一的,。。

    首先对所有点排序。。然后我们可以考虑枚举两个矩形的分割线。。又因为分界线可以是横的也可以是竖的...所以要枚举两次

    具体可以看代码

    #include<algorithm>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #define N 50015
    #define down(i,r,l) for(int i=r;i>=l;i--)
    #define rep(i,l,r) for(int i=l;i<=r;i++)
    using namespace std;
    typedef long long ll;
    int l[N],r[N],l1[N],r1[N],n;
    ll zs,ans;
    struct node{int x,y;}s[N];
    bool cmp(node x,node y) {return x.x==y.x?x.y<y.y:x.x<y.x;}
    void getans() { 
         ll now;
         sort(s+1,s+1+n,cmp);
         memset(l,60,(n+10)<<2); memset(l1,60,(n+10)<<2); memset(r,0,(n+10)<<2); memset(r1,0,(n+10)<<2);
         rep(i,1,n) l[i]=min(l[i-1],s[i].y),r[i]=max(r[i-1],s[i].y);
         down(i,n,1) l1[i]=min(l1[i+1],s[i].y),r1[i]=max(r1[i+1],s[i].y);
         rep(i,2,n) {
              now=(ll)(s[i-1].x-s[1].x) * (ll)(r[i-1]-l[i-1]) + (ll)(s[n].x-s[i].x) * (ll)(r1[i]-l1[i]); 
              ans=min(ans,now);
         }
    }
    int main () {
         int miny,minx,maxx,maxy;
         miny=minx=1000000050,maxy=maxx=0;
         cin>>n; rep(i,1,n) {
             scanf("%d%d",&s[i].x,&s[i].y),miny=min(miny,s[i].y),maxy=max(maxy,s[i].y);;
             maxx=max(maxx,s[i].x); minx=min(minx,s[i].x);
         }
         ans=(ll)(maxy-miny)*(ll)(maxx-minx),zs=ans;
         getans();
         rep(i,1,n) swap(s[i].x,s[i].y);
         getans(); 
         printf("%lld
    ",zs-ans);
    }
    View Code
  • 相关阅读:
    解决Tomcat无法shutdown进程
    ConcurrentHashMap Put()操作示例代码
    Spring Data JPA
    JAVA CAS原理
    多线程
    多线程
    java 虚拟机--新生代与老年代GC
    TCP协议中的三次握手和四次挥手
    java中volatile关键字
    Java多线程
  • 原文地址:https://www.cnblogs.com/Bloodline/p/5492311.html
Copyright © 2011-2022 走看看