CodeForcesGym 100524A Astronomy Problem

Astronomy Problem

Time Limit: 8000ms

Memory Limit: 524288KB

This problem will be judged on CodeForcesGym. Original ID: 100524A
64-bit integer IO format: %I64d      Java class name: (Any)

 

解题：暴力搞

$设两个点的坐标分别为(x_1,y_1),(x_2,y_2)且有x_1^2+y_1^2 = a^2,x_2^2+y_2^2=b^2,那么有(x_1-x_2)^2+(y_1-y_2)^2=c^2。$

$于是推出一个错误的公式，a^2+b^2-c^2=2\times (x_1\times x_2+y_1\times y_2),初略的想法是：等式右边是固定的，所以可以根据这个进行排序，左边是不定的$

$可以用左边找右边，可是，可是，WA第7组数据. 因为，你算多了，而且这个等式本身就不对$

$但是，但是，这个错误的等式可以为我们省出不少的点对，只要不满足这个等式的点对，一定不是符合条件的，但是满足这个等式，也有可能不符合条件$

$所以只会找出这些点对，进行一一检验即可$

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3010;
struct Point{
    LL val,a,b,c;
    Point(LL val = 0,LL a = 0,LL b = 0,LL c = 0){
        this->val = val;
        this->a = a;
        this->b = b;
        this->c = c;
    }
    bool operator<(const Point &rhs)const{
        return val < rhs.val;
    }
}P[maxn*maxn];
int x[maxn],y[maxn],n,q,tot;
LL calc(int a,int b){
    LL tmp = (LL)(x[a] - x[b])*(x[a] - x[b]);
    return tmp + (LL)(y[a] - y[b])*(y[a] - y[b]);
}
int main(){
    freopen("astronomy.in","r",stdin);
    freopen("astronomy.out","w",stdout);
    while(~scanf("%d",&n)){
        if(!n) break;
        tot = 0;
        for(int i = 1; i <= n; ++i){
            scanf("%d%d",x + i,y + i);
            for(int j = 1; j < i; ++j){
                LL tmp = ((LL)x[i]*x[j] + (LL)y[i]*y[j])<<1;
                P[tot++] = Point(tmp,calc(j,0),calc(i,0),calc(i,j));
            }
        }
        sort(P,P + tot);
        scanf("%d",&q);
        while(q--){
            LL a,b,c;
            scanf("%I64d%I64d%I64d",&a,&b,&c);
            LL tmp = a - c + b;
            if(n == 1 || tmp&1){
                puts("0");
                continue;
            }
            int low = lower_bound(P,P + tot,Point(tmp,0,0,0)) - P,ret = 0;
            while(low < tot && P[low].val == tmp){
                if(P[low].a == a && P[low].b == b || P[low].a == b && P[low].b == a) ++ret;
                ++low;
            }
            printf("%d\n",ret);
        }
    }
    return 0;
}
/*
4
0 2
2 0
4 0
0 -4
2
4 16 20
16 4 20
*/

 

然后学习了下某位大神的写法

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3010;
unordered_map<LL,vector<int>>ump;
unordered_map<LL,int>ret;
map<pair<LL,LL>,vector<pair<LL,int>>>Q;
struct Point {
    LL x,y;
    Point(LL x = 0,LL y = 0) {
        this->x = x;
        this->y = y;
    }
} p[maxn];
LL calc(const Point &a,const Point &b) {
    LL tmp = (a.x - b.x)*(a.x - b.x);
    return tmp + (a.y - b.y)*(a.y - b.y);
}
set<LL>st;
int ans[maxn];
int main() {
#define NAME "astronomy"
    freopen(NAME".in","r",stdin);
    freopen(NAME".out","w",stdout);
    int n,q;
    while(~scanf("%d",&n)) {
        if(!n) break;
        ump.clear();
        Q.clear();
        memset(ans,0,sizeof ans);
        for(int i = 1; i <= n; ++i) {
            scanf("%I64d%I64d",&p[i].x,&p[i].y);
            ump[p[i].x*p[i].x+p[i].y*p[i].y].push_back(i);
        }
        scanf("%d",&q);
        for(int i = 0; i < q; ++i){
            LL a,b,c;
            scanf("%I64d%I64d%I64d",&a,&b,&c);
            Q[make_pair(a,b)].push_back(make_pair(c,i));
        }
        for(auto &it:Q){
            st.clear();
            for(auto &it2:it.second) st.insert(it2.first);
            LL a = it.first.first,b = it.first.second;
            ret.clear();
            for(auto &x:ump[a]){
                for(auto &y:ump[b]){
                    if(a == b && x < y) continue;
                    if(x == y) continue;
                    LL d = calc(p[x],p[y]);
                    if(st.find(d) == st.end()) continue;
                    ret[d]++;
                }
            }
            for(auto &v:it.second)
                ans[v.second] = ret[v.first];
        }
        for(int i = 0; i < q; ++i)
            printf("%d\n",ans[i]);
    }
    return 0;
}
/*
4
0 2
2 0
4 0
0 -4
2
4 16 20
16 4 20
*/