UVa 10245 The Closest Pair Problem (分治)

题意：给定 n 个点，求最近两个点的距离。

析：直接求肯定要超时的，利用分治法，先把点分成两大类，答案要么在左边，要么在右边，要么一个点在左边一个点在右边，然后在左边或右边的好求，那么对于一个在左边一个在右边的，我们可以先求全在左边或右边的最小值，假设是d，那么一个点在左边，一个点在右边，那么横坐标之差肯定小于d，才能替换d，同样的纵坐标也是，并且这样的点并不多，然后就可以先选出来，再枚举。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}

struct Point{
double x, y;
bool operator < (const Point &p) const{
return x < p.x || x == p.x && y < p.y;
}
};
Point a[maxn];

bool cmp(const Point &lhs, const Point &rhs){
return lhs.y < rhs.y;
}

double dfs(Point *a, int n){
if(n < 2)  return inf;
int m = n / 2;
double mid = a[m].x;
double d = min(dfs(a, m), dfs(a+m, n-m));
vector<Point> y;
for(int i = 0; i < n; ++i)
if(fabs(mid - a[i].x) < d)  y.push_back(a[i]);

sort(y.begin(), y.end(), cmp);
for(int i = 0; i < y.size(); ++i)
for(int j = i+1; j < y.size(); ++j){
if(fabs(y[i].y-y[j].y) >= d)  continue;
double xx = y[i].x - y[j].x;
double yy = y[i].y - y[j].y;
d = min(d, sqrt(xx*xx + yy*yy));
}
return d;
}

int main(){
while(scanf("%d", &n) == 1 && n){
for(int i = 0; i < n; ++i)  scanf("%lf %lf", &a[i].x, &a[i].y);
sort(a, a + n);
double ans = dfs(a, n);
if(ans >= 10000.0)  printf("INFINITY\n");
else  printf("%.4f\n", ans);
}
return 0;
}