ural 1245. Pictures

1245. Pictures

Time limit: 1.0 second
Memory limit: 64 MB
Artist Ivanov (not the famous Ivanov who painted "Christ's apparition to people", but one of the many namesakes) once managed to rent inexpensively an excellent studio. Alas, as he soon discovered, the inexpensiveness was caused by objective reasons. A murder happened long ago in the house where he rented the room, and now the ghost living in the house each night renews blood spots on the walls of all the rooms. Ivanov's studio did not escape this damnation.
Nevertheless, being a creative person, Ivanov quickly found a simple solution to the problem. He decided to paint one or two pictures and hang them on the (single) wall where the spots appear each night so that the spots would be covered by the pictures. Of course, he does not want to spend too much time doing this work. That is why he plans to use not more than two pictures and wants the total area of the pictures to be minimal.
All the blood spots are circles. Each picture has a rectangular form with sides parallel to the axes, and the minimally possible size of a picture in each of the dimensions is 100 millimeters. If it is necessary to paint two pictures, then they should be hanged to the wall without overlaying. Each spot must be covered by exactly one picture.

Input

The first line contains the number of the spots N, 0 < N ≤ 1000. Each of the next N lines contains the description of the corresponding spot. A spot is described by three positive integers; they are the radius of the spot and the Cartesian coordinates of the center of the spot. Everything is measured in millimeters and all these numbers do not exceed 10000.

Output

Output the minimal total area (in square millimeters) of the pictures (not more than two) necessary to cover all the spots.

Sample

inputoutput
3
50 50 50
50 250 50
10 150 250
 
40000
Problem Author: Alexander Petrov (text — Leonid Volkov)
Problem Source: Ural State University Personal Programming Contest, March 1, 2003
Difficulty: 898
 
题意：平面上有一些圆，半径ri，圆心（xi，yi），问用不超过两个矩阵覆盖他们的最小面积。注意：一个圆不能被两个矩形覆盖。
分析：显然，因为一个圆不能被两个矩形覆盖，瞬间变的简单。
矩形边界必为某个圆的上下左右的切线。
枚举即可。
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <deque>
  6 #include <vector>
  7 #include <queue>
  8 #include <iostream>
  9 #include <algorithm>
 10 #include <map>
 11 #include <set>
 12 #include <ctime>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 typedef double DB;
 17 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 18 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 19 #define Rep(i, t) for(int i = (0); i < (t); i++)
 20 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 21 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 22 #define MIT (2147483647)
 23 #define INF (1000000001)
 24 #define MLL (1000000000000000001LL)
 25 #define sz(x) ((int) (x).size())
 26 #define clr(x, y) memset(x, y, sizeof(x))
 27 #define puf push_front
 28 #define pub push_back
 29 #define pof pop_front
 30 #define pob pop_back
 31 #define ft first
 32 #define sd second
 33 #define mk make_pair
 34 inline void SetIO(string Name)
 35 {
 36     string Input = Name+".in",
 37     Output = Name+".out";
 38     freopen(Input.c_str(), "r", stdin),
 39     freopen(Output.c_str(), "w", stdout);
 40 }
 41 
 42 inline int Getint()
 43 {
 44     int Ret = 0;
 45     char Ch = ' ';
 46     bool Flag = 0;
 47     while(!(Ch >= '0' && Ch <= '9'))
 48     {
 49         if(Ch == '-') Flag ^= 1;
 50         Ch = getchar();
 51     }
 52     while(Ch >= '0' && Ch <= '9')
 53     {
 54         Ret = Ret * 10 + Ch - '0';
 55         Ch = getchar();
 56     }
 57     return Flag ? -Ret : Ret;
 58 }
 59 
 60 const int N = 1010, M = 100;
 61 struct Point
 62 {
 63     int x, y, r;
 64     
 65     inline void Read()
 66     {
 67         r = Getint();
 68         x = Getint();
 69         y = Getint();
 70     }
 71     
 72     inline bool operator <(const Point &A) const
 73     {
 74         return x < A.x;
 75     }
 76 } Arr[N];
 77 int n;
 78 int LU[N], LD[N], RU[N], RD[N], Left[N], Right[N];
 79 int Ans = MIT;
 80 
 81 inline void Input() 
 82 {
 83     n = Getint();
 84     For(i, 1, n) Arr[i].Read();
 85 }
 86 
 87 inline void Work()
 88 {
 89     sort(Arr + 1, Arr + 1 + n);
 90     Right[0] = -INF, LD[0] = INF, LU[0] = -INF;
 91     For(i, 1, n)
 92     {
 93         Right[i] = max(Right[i - 1], Arr[i].x + Arr[i].r);
 94         LU[i] = max(LU[i - 1], Arr[i].y + Arr[i].r);
 95         LD[i] = min(LD[i - 1], Arr[i].y - Arr[i].r);
 96     }
 97     Left[n + 1] = INF, RD[n + 1] = INF, RU[n + 1] = -INF;
 98     Ford(i, n, 1)
 99     {
100         Left[i] = min(Left[i + 1], Arr[i].x - Arr[i].r);
101         RU[i] = max(RU[i + 1], Arr[i].y + Arr[i].r);
102         RD[i] = min(RD[i + 1], Arr[i].y - Arr[i].r);
103     }
104     
105     For(i, 2, n)
106         if(Right[i - 1] <= Left[i])
107             Ans = min(Ans, 
108             max(M, Right[i - 1] - Left[1]) * max(M, LU[i - 1] - LD[i - 1]) +
109             max(M, Right[n] - Left[i]) * max(M, RU[i] - RD[i]));
110 }
111 
112 inline void Solve()
113 {
114     Work();
115     For(i, 1, n) swap(Arr[i].x, Arr[i].y);
116     Work();
117     
118     Ans = min(Ans, max(M, Right[n] - Left[1]) * max(M, LU[n] - LD[n]));
119     
120     printf("%d\n", Ans);
121 }
122 
123 int main()
124 {
125     #ifndef ONLINE_JUDGE 
126     SetIO("D");
127     #endif 
128     Input();
129     Solve();
130     return 0;
131 }
View Code

 

posted @ 2015-11-14 16:06  yanzx6  阅读(239)  评论(0)  编辑  收藏  举报