「SCOI2015」小凸玩矩阵

问题分析

题目给了充足的暗示，我们只需要二分答案然后跑匈牙利即可。要相信匈牙利的速度

参考程序

#include <bits/stdc++.h>

using namespace std;

const int Maxn = 310;

const int INF = 2147483647;

int N, M, K, A[ Maxn ][ Maxn ], Max, Min;

int Map[ Maxn ][ Maxn ], Vis[ Maxn ], Mx[ Maxn ], My[ Maxn ];

int Dfs( int x ) {

	for( int i = 1; i <= M; ++i ) {

		if( Map[ x ][ i ] == 0 ) continue;

		if( Vis[ i ] ) continue;

		Vis[ i ] = 1;

		if( My[ i ] == 0 || Dfs( My[ i ] ) ) {

			Mx[ x ] = i; My[ i ] = x;

			return 1;

		}

	}

	return 0;

}

bool Check( int Mid ) {

	memset( Map, 0, sizeof( Map ) );

	memset( Mx, 0, sizeof( Mx ) );

	memset( My, 0, sizeof( My ) );

	for( int i = 1; i <= N; ++i )

		for( int j = 1; j <= M; ++j )

			if( A[ i ][ j ] <= Mid ) Map[ i ][ j ] = 1;

	int Ans = 0;

	for( int i = 1; i <= N; ++i ) {

		memset( Vis, 0, sizeof( Vis ) );

		Ans += Dfs( i );

	}

	return Ans >= ( N - K + 1 );

}

int main() {

	scanf( "%d%d%d", &N, &M, &K );

	for( int i = 1; i <= N; ++i )

		for( int j = 1; j <= M; ++j )

			scanf( "%d", &A[ i ][ j ] );

	Max = 0, Min = INF;

	for( int i = 1; i <= N; ++i )

		for( int j = 1; j <= M; ++j )

			Max = max( A[ i ][ j ], Max ), Min = min( A[ i ][ j ], Min );

	int Ans = 0;

	while( Max >= Min ) {

		int Mid = ( Max + Min ) >> 1;

		if( Check( Mid ) ) Ans = Mid, Max = Mid - 1; else Min = Mid + 1;

	}

	printf( "%d\n", Ans );

	return 0;

}

巴特西

「SCOI2015」小凸玩矩阵

问题分析

参考程序

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