UVA 1592 Database

题目链接：https://vjudge.net/problem/UVA-1592

题目翻译摘自《算法禁赛入门经典》

题目大意

　　输入一个 n 行 m 列的数据库（1 ≤ n ≤ 10000，1 ≤ i ≤ 10），是否存在两个不同行 r₁，r₂ 和两个不同列 c₁，c₂，使得每列的两个单元格相同（即（r₁，c₁）和（r₂，c₁）相同，（r₁，c₂）和 （r₂，c₂）相同）。

分析

　　暴力枚举复杂度为$O(n^2 * m^2)$，会超时，这里我们可以只枚举列，然后把每一行对应的两个单元格放到有序表里，如果有重复，就说明找到了，时间复杂度为$O(n * \log n * m^2)$。

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 << "]" << "\n";
    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< SI > VSI;
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-10;
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;

struct Line{
    int id;
    VI a;
    
    Line() {}
    Line(int x) : id(x) {}
    
    int operator[] (const int x) const {
        return a[x];
    }
};

int n, m;
MSI str_id;
int len;
string str;

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    while(~scanf("%d %d\n", &n, &m)) {
        vector< Line > vL;
        bool ans = true;
        
        For(i, 1, n) {
            Line t(i);
            
            For(j, 1, m) {
                getline(cin, str, ",\n"[j == m]);
                if(str_id.find(str) == str_id.end()) {
                    str_id[str] = ++len;
                    t.a.PB(len);
                }
                else t.a.PB(str_id[str]);
            }
            vL.PB(t);
        }
        
        PII r, c;
        // 枚举列，保存成映射 
        For(i, 1, m) {
            For(j, i + 1, m) {
                MPIII mpii;
                For(k, 1, n) {
                    PII tmp = MP(vL[k - 1][i - 1], vL[k - 1][j - 1]);
                    if(mpii.find(tmp) != mpii.end()) {
                        r = MP(mpii[tmp], k);
                        c = MP(i, j);
                        ans = false;
                        goto Dark_Souls;
                    }
                    else mpii[tmp] = k;
                }
            }
        }
        
        Dark_Souls: {
            if(ans) cout << "YES" << endl;
            else {
                cout << "NO" << endl;
                cout << r.ft << " " << r.sd << endl;
                cout << c.ft << " " << c.sd << endl;
            }
        }
    }
    return 0;
}