C 解法, 执行用时: 3ms, 内存消耗: 376KB, 提交时间: 2020-11-14

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int sort(long* list, int len, int* arrangedindex){
 //对list中的所有数使用归并排序方法从大到小排列，将这些数对应在list中的下标存储至arrangedindex
    if(len <= 0)return 0;
    if(len == 1){
        arrangedindex[0] = 0;
        return 0;
    }
    int middle = len/2;
    long* newlist = (long*)(list + middle);
    int* arrng1 = (int*)malloc(middle*sizeof(int));
    int* arrng2 = (int*)malloc((len - middle)*sizeof(int));
    sort(list, middle, arrng1);
    sort(newlist, (len-middle), arrng2);
    int index = 0;
    int a,b;
    a = 0;
    b = 0;
    while(index < len){
        if(a >= middle){
            while(b < (len - middle)){
                arrangedindex[index] = arrng2[b] + middle;
                b++;
                index++;
            }
            break;
        }
        if(b >= (len - middle)){
            while(a < middle){
                arrangedindex[index] = arrng2[a];
                a++;
                index++;
            }
            break;
        }
        if(list[arrng1[a]] <= newlist[arrng2[b]]){
            arrangedindex[index] = arrng2[b] + middle;
            b++;
            index++;
            continue;
        }else{
            arrangedindex[index] = arrng1[a];
            a++;
            index++;
            continue;
        }
    }
    free(arrng1);
    free(arrng2);
    return 0;
}

int main(){
    int n,x1,y1,x2,y2;
    while(scanf("%d %d %d %d %d", &n, &x1, &y1, &x2, &y2) != EOF){
        int i,j,k;
        int x,y;
        long* ftos1 = (long*)malloc(n*sizeof(long));//distence's square from floower to spring 1
        long* ftos2 = (long*)malloc(n*sizeof(long));//distence's square from floower to spring 2
        for(i = 0; i < n; i++){
            scanf("%d %d", &x, &y);
            ftos1[i] = ((long)(x - x1))*((long)(x - x1)) + ((long)(y - y1))*((long)(y - y1));
            ftos2[i] = ((long)(x - x2))*((long)(x - x2)) + ((long)(y - y2))*((long)(y - y2));
        }
        int* arrangedindex = (int*)malloc(n*sizeof(int));
        sort(ftos1, n, arrangedindex);
        long min = ftos1[arrangedindex[0]];
        long maxof2 = -1;
        for(i = 0; i < n - 1; i++){
            j = arrangedindex[i];
            if(ftos2[j] > maxof2){
                maxof2 = ftos2[j];
            }
            k = arrangedindex[i+1];
            if((ftos1[k] + maxof2) < min){
                min = ftos1[k] + maxof2;
            }
        }
        if(ftos2[arrangedindex[n-1]] > maxof2){
            maxof2 = ftos2[arrangedindex[n-1]];
        }
        if(maxof2 < min){
            min = maxof2;
        }
        printf("%ld\n", min);
        free(arrangedindex);
        free(ftos1);
        free(ftos2);
    }
    return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 376KB, 提交时间: 2020-08-02

#include<bits/stdc++.h>
using namespace std;
int main() {
    int n, x1, y1, x2, y2;
    cin >> n >> x1 >> y1 >> x2 >> y2;
    vector<pair<long, long>> v;
    for (long i = 0; i < n; i++) {
        int xf, yf;
        cin >> xf >> yf;
        v.push_back(pair<long, long>(pow(xf - x1, 2) + pow(yf - y1, 2),
                                     pow(xf - x2, 2) + pow(yf - y2, 2)));
    }
    sort(v.begin(), v.end(),
        [](pair<long, long>& p1, pair<long, long>& p2) {return p1.first < p2.first;});
    long ans = v[n - 1].first, jMax = v[n-1].second;
    for (int i = n - 2; i >= 0; i--) {
        ans = min(ans, (v[i].first + jMax));
        jMax = max(jMax, v[i].second);
    }
    cout<<min(ans,jMax); 
    return 0;
}