UVA 201 Squares

题目链接：https://vjudge.net/problem/UVA-201

题目翻译摘自《算法禁赛入门经典》

题目大意

　　有 n 行 n 列（2 ≤ n ≤ 9）的小黑点，还有 m 条线段连接其中的一些黑点。统计这些线段连成 了多少个正方形（每种边长分别统计）。 行从上到下编号为 1～n，列从左到右编号为 1～n。边用 H i j 和 V i j 表示，分别代表边 (i, j)-(i, j + 1) 和 (i, j)-(i + 1, j)。

分析

　　直接暴力时间复杂度是 O(n⁴)，可以扩充矩阵，然后用前缀和优化到 O(n³)。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< VI > VVI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< int, string > MIS;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef map< string, string > MSS;
typedef map< PII, string > MPIIS;
typedef map< PII, PII > MPIIPII;
typedef multimap< int, int > MMII;
typedef multimap< string, int > MMSI;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-6;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e4 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int T, n, m; 
int A[23][23], rowPreSum[23][23], colPreSum[23][23];
MII sqr; // size->cnt

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    while(cin >> n >> m) {
        ms0(A);
        ms0(rowPreSum);
        ms0(colPreSum);
        sqr.clear();
        n = 2 * n - 1;
        Rep(i, m) {
            string str;
            int x, y;
            cin >> str >> x >> y;
            x = 2 * x - 1;
            y = 2 * y - 1;
            if(str == "H") For(j, y, y + 2) A[x][j] = 1;
            else For(j, y, y + 2) A[j][x] = 1;
        }
        // 计算矩阵的行列前缀和 
        For(i, 1, n) {
            For(j, 1, n) {
                rowPreSum[i][j] += rowPreSum[i][j - 1] + A[i][j];
                colPreSum[i][j] += colPreSum[i][j - 1] + A[j][i];
            }
        }
        
        // 枚举所有正方形对角点对 
        For(i, 1, n) {
            For(j, 1, n) {
                For(k, 2, n) {
                    if(i + k > n || j + k > n) break;
                    if(rowPreSum[i][j + k] - rowPreSum[i][j - 1] == k + 1 &&
                        rowPreSum[i][j + k] - rowPreSum[i][j - 1] == colPreSum[j][i + k] - colPreSum[j][i - 1] &&
                        colPreSum[j][i + k] - colPreSum[j][i - 1] == rowPreSum[i + k][j + k] - rowPreSum[i + k][j - 1] &&
                        rowPreSum[i + k][j + k] - rowPreSum[i + k][j - 1] == colPreSum[j + k][i + k] - colPreSum[j + k][i - 1]
                        ) ++sqr[(k + 1) / 2];
                }
            }
        }
        
        if(T++) printf("
**********************************

");
        printf("Problem #%d

", T);
        if(sqr.empty()) printf("No completed squares can be found.
");
        else foreach(i, sqr) printf("%d square (s) of size %d
", i->sd, i->ft);
    }
    return 0;
}