CodeForces 266CBelow the Diagonal

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2016暑期集训1-D

CodeForces 266C Below the Diagonal

想法题，矩阵换行换列

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传送门：CodeForce

题意

给你一个n阶01矩阵，保证里面有n-1个1，其余0。你每次可以交换其中两行或两列，不要求交换次数最小，但不能大于10000。最后使得矩阵的1全在主对角线下方。

思路

从n阶开始看，每次先使第n列全0，再使第n行有1，然后递归处理n-1阶矩阵，直到n=1。
对矩阵加边，记录每行每列的1的个数，递归之前处理边，如果第n行某位置有1，那么把该列的1个数减1。
记录交换方法，最后输出。

代码

#include <iostream>#include<cstdio>#include<algorithm>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<queue>#include<stack>using namespace std;const int MAXN = 1007;const int oo = 2000000007;const long long int loo = 2000000000000000007ll;typedef pair<int , pair<int , int > > P;queue <P> que;int q [ MAXN ] [ MAXN ];void solve ( int n ){    if ( n == 1 ) return;    else    {        if ( q [ 0 ] [ n ] == 0 )//最后一列全是0        {            if ( q [ n ] [ 0 ] != 0 )//最后一行有1            {                for ( int i = 1; i <= n; i++ )                {                    if ( q [ n ] [ i ] == 1 )                    {                        q [ 0 ] [ i ]--;                    }                }                solve ( n - 1 );            }            else//最后一行没有1            {                for ( int i = 1; i <= n; i++ )                {                    if ( q [ i ] [ 0 ] != 0 )                    {                        swap ( q [ i ] , q [ n ] );                        que.push ( P ( 1 , make_pair ( i , n ) ) );                        break;                    }                }                for ( int i = 1; i <= n; i++ )                {                    if ( q [ n ] [ i ] == 1 )                    {                        q [ 0 ] [ i ]--;                    }                }                solve ( n - 1 );            }        }        else//最后一列有1        {            for ( int i = 1; i <= n; i++ )            {                if ( q [ 0 ] [ i ] == 0 )                {                    for ( int j = 0; j <= n; j++ )                    {                        swap ( q [ j ] [ i ] , q [ j ] [ n ] );                    }                    que.push ( P ( 2 , make_pair ( i , n ) ) );                    break;                }            }            if ( q [ n ] [ 0 ] != 0 )//最后一行有1            {                for ( int i = 1; i <= n; i++ )                {                    if ( q [ n ] [ i ] == 1 )                    {                        q [ 0 ] [ i ]--;                    }                }                solve ( n - 1 );            }            else//最后一行没有1            {                for ( int i = 1; i <= n; i++ )                {                    if ( q [ i ] [ 0 ] != 0 )                    {                        swap ( q [ i ] , q [ n ] );                        que.push ( P ( 1 , make_pair ( i , n ) ) );                        break;                    }                }                for ( int i = 1; i <= n; i++ )                {                    if ( q [ n ] [ i ] == 1 )                    {                        q [ 0 ] [ i ]--;                    }                }                solve ( n - 1 );            }        }    }}int main ( ){    while ( que.size ( ) != 0 )    {        que.pop ( );    }    memset ( q , 0 , sizeof ( q ) );    int n;    scanf ( "%d" , &n );    int a , b;    while ( scanf ( "%d%d" , &a , &b ) == 2 )    {        q [ a ] [ b ] = 1;        q [ 0 ] [ b ]++;        q [ a ] [ 0 ]++;    }    solve ( n );    printf ( "%d\n" , que.size ( ) );    while ( que.size ( ) != 0 )    {        P p = que.front ( );        que.pop ( );        printf ( "%d %d %d\n" , p.first , p.second.first , p.second.second );    }    system ( "pause" );    return 0;}

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