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 = ;
 struct Point{
     LL val,a,b,c;
     Point(LL val = ,LL a = ,LL b = ,LL c = ){
         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 = ;
         for(int i = ; i <= n; ++i){
             scanf("%d%d",x + i,y + i);
             for(int j = ; j < i; ++j){
                 LL tmp = ((LL)x[i]*x[j] + (LL)y[i]*y[j])<<;
                 P[tot++] = Point(tmp,calc(j,),calc(i,),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 ==  || tmp&){
                 puts("");
                 continue;
             }
             int low = lower_bound(P,P + tot,Point(tmp,,,)) - P,ret = ;
             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 ;
 }
 /*
 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 = ;
 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 = ,LL y = ) {
         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,,sizeof ans);
         for(int i = ; 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 = ; 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 = ; i < q; ++i)
             printf("%d\n",ans[i]);
     }
     return ;
 }
 /*
 4
 0 2
 2 0
 4 0
 0 -4
 2
 4 16 20
 16 4 20
 */