UVa1592 数据库（摘）

输入一个n行m列的数据库（1<=n<=10000,1<=m<=10),是否存在两个不同行r1，r2和两个不同列c1，c2，使得这两行和这两行相同（即（r1，c1）和（r2，c1）相同，

（r1，c2）和（r2，c2）相同）。例如，对于如图所示的数据库，第2、3行和第2、3列满足要求

 

How to compete in AM ICPC	Peter	peter@neerc.ifmo.ru
How to win in AM ICPC	Michael	michael@neerc.ifmo.ru
Notes from AM ICPC champion	Michael	michael@neerc.ifmo.ru

 

Input

 

Input contains several datasets. The first line of each dataset contains two integer numbersn and m (1n10000, 1m10), the number of rows and columns in the table. The following n lines contain table rows. Each row hasm column values separated by commas. Column values consist of ASCII characters from space (ASCII code 32) to tilde (ASCII code 126) with the exception of comma (ASCII code 44). Values are not empty and have no leading and trailing spaces. Each row has at most 80 characters (including separating commas).

 

Output

 

For each dataset, if the table is in PNF write to the output file a single word ``YES" (without quotes). If the table is not in PNF, then write three lines. On the first line write a single word ``NO" (without quotes). On the second line write two integer row numbers r₁ andr₂ (1r₁,r₂n,r₁r₂), on the third line write two integer column numbers c₁ andc₂ (1c₁,c₂m,c1c2), so that values in columnsc₁ andc₂ are the same in rowsr₁ andr₂.

 

 Sample input

3 3
How to compete in ACM ICPC,Peter,peter@neerc.ifmo.ru
How to win ACM ICPC,Michael,michael@neerc.ifmo.ru
Notes from ACM ICPC champion,Michael,michael@neerc.ifmo.ru
2 3
1,Peter,peter@neerc.ifmo.ru
2,Michael,michael@neerc.ifmo.ru

 

 Sample output

NO
2 3
2 3
YES

#include<iostream>
#include<cstdio>
#include<map>
#include<string>
#include<vector>
using namespace std;

const int ROW = 10000 + 10;
const int COL = 10 + 5;
int n,m;
map<string, int> IDcache;
vector<string> Strcache;
vector<int> Text[ROW];           //处理后的文本，每个字符串都对应一个编号
struct node
{
    int x,y;
    node(int x, int y):x(x),y(y) { }
    bool operator < (const node& r) const{           //重载'<'
        return x<r.x || x==r.x&&y<r.y; 
    }       
};
map<node,int> data;

int f(string str)
{
    if(IDcache.count(str)) return IDcache[str];
    Strcache.push_back(str);
    return IDcache[str] = Strcache.size()-1;
}

void read()
{
    string str;
    char ch = getchar();
    for(int i=0;i<n;i++)
    {
        for(;;)
        {
            ch = getchar();
            if(ch=='\n'||ch=='\r') {
                if(!str.empty()) Text[i].push_back(f(str));
                str.clear(); 
                break;
            }
            if(ch!=',') str += ch;
            else { 
                Text[i].push_back(f(str)); 
                str.clear();
            }
         }
    }
}

void solve()
{
    int x,y,c1,c2;
    for(c1=0;c1<m;c1++)
    {
        for(c2=c1+1;c2<m;c2++)
        {
            data.clear();
            for(int r=0;r<n;r++)
            {
                x = Text[r][c1]; y = Text[r][c2];
                node p(x,y);
                if(!data.count(p)) data[p] = r;
                else{
                    cout<<"NO"<<endl;
                    cout<<data[p]+1<<" "<<r+1<<endl<<c1+1<<" "<<c2+1<<endl;
                    return;
                }
            }
        }
    }
    cout<<"YES"<<endl;
}

int main()
{
    while(cin>>n>>m)
    {
        read();
        solve();
        for(int i=0;i<n;i++) Text[i].clear();
        IDcache.clear(); Strcache.clear();
    }
//system("pause");
    return 0;
}