BZOJ 4443： 小凸玩矩阵【二分图】

我是传送门

先看题目，从数列中选第K小，很容易想到二分或者单调队列，但这里单调队列显得不是那么合适。而任意两个数不在一行一列，这符合二分图的定义，所以思路就很明了了，找出所有的值然后去二分找答案。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define oo 0x7f7f7f7f
#define get(x) scanf( "%d", &x )
#define put(x) printf( "%d", x )
#define cln(x) memset( x, 0, sizeof(x) )
using namespace std;

const int R = 255;
int f[R];
int p[R];
int sq[R][R];
int head[R];
int to[R*R];
int next[R*R];
int n, m, k, ans, tot;

void add( int x, int y )
{
    tot++;
    to[tot] = y;
    next[tot] = head[x];
    head[x] = tot;
}

int DFS( int x, int t )
{
    for ( int i = head[x]; i != -1; i = next[i] )
    {
        if ( p[to[i]] != t )
        {
            int y = to[i];
            p[y] = t;
            if ( f[y] == 0 || DFS( f[y], t ) )
            {
                f[y] = x;
                return 1;
            }
        }
    }
    return 0;
}

int solve( int l, int r )
{
    if ( l > r ) return  l;
    int mid = ( l + r ) >> 1;
    ans = 0;
    tot = 0;

    for ( int i = 1; i <= n; i++ )
        head[i] = -1;
    for ( int i = 1; i <= m; i++ )
        p[i] = f[i] = 0;
    for ( int i = 1; i <= n; i++ )
        for ( int j = 1; j <= m; j++ )
            if ( sq[i][j] <= mid )
                add( i, j );
    for ( int i = 1; i <= n; i++ )
        ans += DFS( i, i );

    if ( ans >= n - k + 1 )
        return solve(l, mid - 1);
    else
        return solve( mid + 1, r );
}

int main()
{
    get(n), get(m), get(k);
    for ( int i = 1; i <= n; i++ )
        for ( int j = 1; j <= m; j++ )
        {
            get(sq[i][j]);
            ans = max( ans, sq[i][j] );
        }
    put( solve( 1, ans ) );
    return 0;
}

↑除了MLE所有错误类型都被我弄出来了，真是伤不起Orz