LA 3695 Distant Galaxy（扫描线）

Distant Galaxy

Time Limit: 3000ms
Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 3695

64-bit integer IO format: %lld      Java class name: Main

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[PDF Link]

You are observing a distant galaxy using a telescope above the Astronomy Tower, and you think that a rectangle drawn in that galaxy whose edges are parallel to coordinate axes and contain maximum star systems on its edges has a great deal to do with the mysteries
of universe. However you do not have the laptop with you, thus you have written the coordinates of all star systems down on a piece of paper and decide to work out the result later. Can you finish this task?

Input

There are multiple test cases in the input file. Each test case starts with one integer N , (1

N

100) ,
the number of star systems on the telescope. N lines follow,
each line consists of two integers: the X and Y coordinates
of the K -th planet system. The absolute value of any coordinate
is no more than 109 , and you can assume that the planets
are arbitrarily distributed in the universe.

N = 0 indicates the end of input file and should not be processed
by your program.

Output

For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.

Sample Input

10
2 3
9 2
7 4
3 4
5 7
1 5
10 4
10 6
11 4
4 6
0

Sample Output

Case 1: 7

Source

Regionals 2006, Asia - Shanghai

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给出n个点坐标，找一个矩形，使边界上的点最多。思路枚举上下边界，然后用扫描线，大白P53

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 10;
struct Node {
int x,y;
bool operator < (const Node & rhs) const {
return x < rhs.x;
}
}p[maxn];
int y[maxn];
int L[maxn],on[maxn],on2[maxn];
int solve(int n)
{
sort(p,p+n);
sort(y,y+n);

int m = unique(y,y+n) - y;
if(m < 3) return n; ///不超过两个y坐标
int ans = 0;
for(int a = 0; a < m; ++a) {
for(int b = a+1; b < m; ++b) {
int k = 0,ymin = y[a],ymax = y[b];
for(int i = 0; i < n; ++i) {
if(i==0 || p[i].x != p[i-1].x) { ///新的竖线
++k;
on[k] = 0;
on2[k] = 0;
L[k] = L[k-1] + on2[k-1] - on[k-1];
}
if(p[i].y > ymin && p[i].y < ymax) ++on[k];
if(p[i].y >= ymin && p[i].y <= ymax) ++on2[k];
}
if(k < 3) return n; ///不超过两个x坐标
int M = 0;
for(int i = 1; i <= k; ++i) {
ans = max(ans,L[i]+on2[i]+M);
M = max(M,on[i]-L[i]);
}
}
}
return ans;
}
int main()
{
int n,cas = 0;
while(scanf("%d",&n)==1&& n) {
++cas;
for(int i = 0; i < n; ++i) {
scanf("%d%d",&p[i].x,&p[i].y);
y[i] = p[i].y;
}
int ans = solve(n);
printf("Case %d: %d\n",cas,ans);
}
return 0;
}