HDU1007 --- Job title

Server collapse, not data backup, back garden blog

HDU1007

Meaning of the questions:
to give you some points on a coordinate plane, let you figure out this pile of points, half of the number of shortest distance two points, the distance?

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Solution:
data \ (the n-6 = 10 ^ \) , if we enumerate each point, then, it is clear that the complexity will reach \ (the n-^ 2 \) , this will certainly timeout.

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Consider using divide and conquer.

1. First, we follow the point \ (x \) axis coordinates from small to large
2. Select a midpoint \ (MID \) , all the current is divided into right and left points \ ((1, mid), (mid + 1, n) \)
   • The results can now be launched only three cases
     • Two points are on the left
     • Two points are on the right
     • One on the left, one on the right
3. For results \ (1 \) and the result \ (2 \) we can find a recursive, recursion boundary when the current number of points is \ (2 \) or (3 \) \ , we can directly calculate the results .
   We have got so \ (Lmin \) and \ (Rmin \) .

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Next, processing the third result .

As we previously come to \ (Lmin \) and \ (Rmin \) , thereby obtaining \ (DIS = min (Lmin, Rmin) \) .

If you want to verify the results of the third, it must be connected by a point to the right of the left point, and the shortest distance, so we naturally choose \ (mid \) points.

Next, at present all points selected and \ (MID \) point \ (X \) axis distance is no greater than \ (DIS \) in all points. (If even \ (X \) axis is greater than the distance dis, count \ (Y \) axis distance farther only)

The chosen point by \ (\ Y) axis coordinates from small to large, and these points can be enumerated.

Intermediate may further prune, if \ (V \) from the point has arrived greater than \ (DIS \) , then the next point can be replaced.

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Additional knowledge (are interested can learn)

In just the last step of the enumeration, in Introduction to Algorithms we have explained.

We will click on it, these points are in a natural \ (X \) axis distance \ (2 * dis \) of a region.

Select a point as a base point , and the distance is less than his (DIS \) \ other points, natural \ (Y \) axis distance is not greater than \ (DIS \)
So we for each point can draw a region, and this both sides in the region up to 4 points, the sum of 8 points.

Because if any of you around while more than a point, does not meet before obtaining \ (Lmin \) .

That is at best only an enumeration of each point eight times.

#include <bits/stdc++.h>
//freopen("in.txt", "r", stdin);

using namespace std;
typedef double dou;
typedef long long ll;
typedef pair<ll, ll> pii;

#define M 200050
#define inf 0x3f3f3f3f
#define mod 998244353
#define W(a) while(a)
#define lowbit(a) a&(-a)
#define left k<<1
#define right k<<1|1
#define ms(a,b) memset(a,b,sizeof(a))
#define debug(a) cout<<#a<<" == "<<a<<endl
#define false_stdio ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

int n;
struct Data {
    dou x;
    dou y;
}node[M];

bool cmpx(Data &a, Data &b) { return a.x < b.x; }
bool cmpy(Data &a, Data &b) { return a.y < b.y; }

dou dis(Data& a, Data &b) { return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y)); }

dou solve(int L, int R) {
    if (L + 1 == R)return dis(node[L], node[R]);//如果目前只有两个点
    if (L + 2 == R)return min(dis(node[L], node[L + 1]), min(dis(node[L], node[R]), dis(node[L + 1], node[R])));//如果目前只有三个点
    int mid = L + R >> 1;
    dou ans = min(solve(L, mid), solve(mid + 1, R));//递归求左右两边最小距离

    //选点
    vector<Data>tmp;
    for (; L <= R; L++) {
        if (fabs(node[mid].x - node[L].x) <= ans) {
            tmp.push_back(node[L]);
        }
    }
    sort(tmp.begin(), tmp.end(), cmpy);//按y坐标排序

    //枚举点
    for (int i = 0; i < tmp.size(); i++) {
        for (int j = i + 1; j < tmp.size(); j++) {
            if (tmp[j].y - tmp[i].y >= ans)break;
            ans = min(ans, dis(tmp[i], tmp[j]));
        }
    }
    return ans;
}

int main() {
    false_stdio;//HDU貌似关闭同步cin还是慢，所以还是用scanf了
    W(scanf("%d", &n) != EOF && n) {
        for (int i = 1; i <= n; i++) scanf("%lf%lf", &node[i].x, &node[i].y);
        sort(node + 1, node + n + 1, cmpx);//按x坐标排序
        printf("%.2lf\n", solve(1, n) / 2.0);
    }
    return 0;
}