Problem A. Nearest Neighbor Search ——BNUOJ 点距离立方体最短

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题目描述：

https://acm.bnu.edu.cn/v3/statments/52296.pdf点击打开链接

求一个点到立方体最短距离平方，如果点在立方体内部则距离为0

思路，求解每个方向的最小距离，然后求其平方和

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 2009#define INF 9999999999999#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}ll my_min(ll x,ll y){ return x>y?y:x;};int main(int argc,char *argv[]){ll x0,y0,z0,x1,y1,z1,x2,y2,z2,x,y,z;    while((scanf("%lld %lld %lld",&x0,&y0,&z0)!=EOF))    {    scanf("%lld %lld %lld",&x1,&y1,&z1);    scanf("%lld %lld %lld",&x2,&y2,&z2);if(x0>=x1&&x0<=x2&&y0>=y1&&y0<=y2&&z0>=z1&&z0<=z2){printf("0\n");continue;}if(x0>=x1&&x0<=x2)x=0;elsex = my_min(abs(x0-x1),abs(x0-x2));if(y0>=y1&&y0<=y2)y=0;elsey = my_min(abs(y0-y1),abs(y0-y2));if(z0>=z1&&z0<=z2)z=0;elsez = my_min(abs(z0-z1),abs(z0-z2));ll re = x*x+y*y+z*z;printf("%lld\n",re);    }    return 0;}

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