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  • POJ 2002 Squares

    题目链接:https://vjudge.net/problem/POJ-2002

    题目大意

      多组数据,每组给定 n 个不同的点,问从这 n 个点中选 4 个,使之能连成正方形,有多少种不同的选法?

    分析

      思路是枚举 2 个点,然后计算出可能的另外两个点,看它们存不存在。
      最优解应该是用哈希表(O(n2))做,不过POJ万年不更新,都不能用 unordered_map 和 unordered_set,我弄了半天还是 wa(菜),最后用二分O(n2logn)过了,时间正好够。

    代码如下

      1 #include <cmath>
      2 #include <ctime>
      3 #include <iostream>
      4 #include <string>
      5 #include <vector>
      6 #include <cstdio>
      7 #include <cstdlib>
      8 #include <cstring>
      9 #include <queue>
     10 #include <map>
     11 #include <set>
     12 #include <algorithm>
     13 #include <cctype>
     14 #include <stack>
     15 #include <deque>
     16 #include <list>
     17 #include <sstream>
     18 using namespace std;
     19  
     20 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
     21 #define Rep(i,n) for (int i = 0; i < (n); ++i)
     22 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
     23 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
     24 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
     25 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
     26 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
     27 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
     28  
     29 #define pr(x) cout << #x << " = " << x << "  "
     30 #define prln(x) cout << #x << " = " << x << endl
     31  
     32 #define LOWBIT(x) ((x)&(-x))
     33  
     34 #define ALL(x) x.begin(),x.end()
     35 #define INS(x) inserter(x,x.begin())
     36 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
     37 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
     38 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
     39 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
     40  
     41 #define ms0(a) memset(a,0,sizeof(a))
     42 #define msI(a) memset(a,inf,sizeof(a))
     43 #define msM(a) memset(a,-1,sizeof(a))
     44 
     45 #define MP make_pair
     46 #define PB push_back
     47 #define ft first
     48 #define sd second
     49  
     50 template<typename T1, typename T2>
     51 istream &operator>>(istream &in, pair<T1, T2> &p) {
     52     in >> p.first >> p.second;
     53     return in;
     54 }
     55  
     56 template<typename T>
     57 istream &operator>>(istream &in, vector<T> &v) {
     58     for (auto &x: v)
     59         in >> x;
     60     return in;
     61 }
     62  
     63 template<typename T1, typename T2>
     64 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     65     out << "[" << p.first << ", " << p.second << "]" << "
    ";
     66     return out;
     67 }
     68 
     69 inline int gc(){
     70     static const int BUF = 1e7;
     71     static char buf[BUF], *bg = buf + BUF, *ed = bg;
     72     
     73     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
     74     return *bg++;
     75 } 
     76 
     77 inline int ri(){
     78     int x = 0, f = 1, c = gc();
     79     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
     80     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
     81     return x*f;
     82 }
     83 
     84 template<class T>
     85 inline string toString(T x) {
     86     ostringstream sout;
     87     sout << x;
     88     return sout.str();
     89 }
     90 
     91 inline int toInt(string s) {
     92     int v;
     93     istringstream sin(s);
     94     sin >> v;
     95     return v;
     96 }
     97 
     98 //min <= aim <= max
     99 template<typename T>
    100 inline bool BETWEEN(const T aim, const T min, const T max) {
    101     return min <= aim && aim <= max;
    102 }
    103  
    104 typedef long long LL;
    105 typedef unsigned long long uLL;
    106 typedef pair< double, double > PDD;
    107 typedef pair< int, int > PII;
    108 typedef pair< int, PII > PIPII;
    109 typedef pair< string, int > PSI;
    110 typedef pair< int, PSI > PIPSI;
    111 typedef set< int > SI;
    112 typedef set< PII > SPII;
    113 typedef vector< int > VI;
    114 typedef vector< double > VD;
    115 typedef vector< VI > VVI;
    116 typedef vector< SI > VSI;
    117 typedef vector< PII > VPII;
    118 typedef map< int, int > MII;
    119 typedef map< LL, int > MLLI;
    120 typedef map< int, string > MIS;
    121 typedef map< int, PII > MIPII;
    122 typedef map< PII, int > MPIII;
    123 typedef map< string, int > MSI;
    124 typedef map< string, string > MSS;
    125 typedef map< PII, string > MPIIS;
    126 typedef map< PII, PII > MPIIPII;
    127 typedef multimap< int, int > MMII;
    128 typedef multimap< string, int > MMSI;
    129 //typedef unordered_map< int, int > uMII;
    130 typedef pair< LL, LL > PLL;
    131 typedef vector< LL > VL;
    132 typedef vector< VL > VVL;
    133 typedef priority_queue< int > PQIMax;
    134 typedef priority_queue< int, VI, greater< int > > PQIMin;
    135 const double EPS = 1e-8;
    136 const LL inf = 0x3fffffff;
    137 const LL infLL = 0x3fffffffffffffffLL;
    138 const LL mod = 1e9 + 7;
    139 const int maxN = 1e6 + 7;
    140 const LL ONE = 1;
    141 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
    142 const LL oddBits = 0x5555555555555555;
    143 
    144 struct Point{
    145     int X, Y;
    146     
    147     inline bool operator< (const Point &x) const {
    148         return X < x.X || X == x.X && Y < x.Y;
    149     }
    150     
    151     inline bool operator== (const Point &x) const {
    152         return X == x.X && Y == x.Y;
    153     }
    154 };
    155 
    156 istream& operator>> (istream& in, Point &x) {
    157     in >> x.X >> x.Y;
    158     return in;
    159 }
    160 
    161 int n, ans; 
    162 Point points[1007];
    163 
    164 int main(){
    165     //freopen("MyOutput.txt","w",stdout);
    166     //freopen("input.txt","r",stdin);
    167     INIT();
    168     while(cin >> n && n) {
    169         ans = 0;
    170         Rep(i, n) cin >> points[i];
    171         sort(points, points + n);
    172         
    173         Rep(i, n) {
    174             For(j, i + 1, n - 1) {
    175                 Point tmpx, tmpy;
    176                 tmpx.X = points[i].X + points[i].Y - points[j].Y;
    177                 tmpx.Y = points[i].Y + points[j].X - points[i].X;
    178                 tmpy.X = points[j].X + points[i].Y - points[j].Y;
    179                 tmpy.Y = points[j].Y + points[j].X - points[i].X;
    180                 
    181                 int x = lower_bound(points, points + n, tmpx) - points;
    182                 if(x >= n || !(points[x] == tmpx)) continue;
    183                 int y = lower_bound(points, points + n, tmpy) - points;
    184                 if(y < n && points[y] == tmpy) ++ans;
    185             }
    186         }
    187         
    188         cout << ans / 2 << endl;
    189     }
    190     return 0;
    191 }
    View Code
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  • 原文地址:https://www.cnblogs.com/zaq19970105/p/11180303.html
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