Solution -「洛谷 P4194」矩阵

$\mathcal{Description}$

  Link.

  给定一个 $n\times m$ 的矩阵 $A$，构造一个 $n\times m$ 的矩阵 $B$，s.t. $(\forall i\in[1,n],j\in[1,m])(b_{ij}\in[L,R])$，且最小化：

$$\max\left\{\max_{i=1}^n\{\left|\sum_{j=1}^m a_{ij}-b_{ij}\right|,\max_{j=1}^m\left| \sum_{i=1}^n a_{ij}-b_{ij} \right|\right\}$$

  输出上式最小值即可。

  $n,m\le200$，$0\le L,R,a_{ij}\le10^3$。

$\mathcal{Solution}$

  不难想到二分答案。记 $r_i=\sum_{j=1}^m a_{ij}$，$c_i=\sum_{j=1}^n a_{ji}$，设当前答案为 $x$，则第 $i$ 行的取值区间为 $[\max\{0,r_i-x\},r_i+x]$，第 $i$ 列的取值区间为 $[\max\{0,c_i-x\},c_i+x]$，然后 $B$ 的每个元素又有限制 $[L,R]$，所以猜测可以通过求上下界可行流来构造 $B$：

• $S$ 连向 $n$ 个行虚点 $r_1,r_2,\cdots,r_n$，流量区间如上；
• 行虚点 $r_i$ 连向第 $i$ 行元素入点 $bi_{ij}$，流量无限制；
• 元素入点 $bi_{ij}$ 连向元素出点 $bo_{ij}$，流量区间 $[L,R]$；
• 元素出点 $bo_{ij}$ 连向列虚点 $c_j$，流量无限制；
• $m$ 个列虚点 $c_1,c_2,\cdots,c_m$ 连向 $T$，流量区间如上。

  求这个有源汇流网络是否存在可行流即可判断 $x$ 是否合法。

  复杂度 $\mathcal O(\log(nR)D)$，其中 $D$ 为这种分层图下 Dinic 算法的复杂度。

$\mathcal{Code}$

/* Clearink */

#include <queue>
#include <cstdio>

const int MAXN = 300, MAXV = 1e3, MAXND = MAXN * ( MAXN + 1 ) * 2 + 2, INF = 0x3f3f3f3f;
int n, m, L, R, rsum[MAXN + 5], csum[MAXN + 5], deg[MAXND + 5];

inline int imin ( const int a, const int b ) { return a < b ? a : b; }
inline int imax ( const int a, const int b ) { return a < b ? b : a; }

struct MaxFlowGraph {
	static const int MAXND = ::MAXND + 2, MAXEG = MAXN * MAXN * 3 + MAXN * 2;
	int ecnt, head[MAXND + 5], S, T, bound, curh[MAXND + 5], d[MAXND + 5];
	struct Edge { int to, flow, nxt; } graph[MAXEG * 2 + 5];

	inline void clear () {
		ecnt = 1;
		for ( int i = 0; i <= bound; ++i ) head[i] = 0;
	}

	inline void link ( const int s, const int t, const int f ) {
		graph[++ecnt] = { t, f, head[s] };
		head[s] = ecnt;
	}

	inline Edge& operator [] ( const int k ) { return graph[k]; }

	inline void operator () ( const int s, const int t, const int f ) {
		#ifdef RYBY
			printf ( "%d %d ", s, t );
			if ( f == INF ) puts ( "INF" );
			else printf ( "%d\n", f );
		#endif
		link ( s, t, f ), link ( t, s, 0 );
	}

	inline bool bfs () {
		static std::queue<int> que;
		for ( int i = 0; i <= bound; ++i ) d[i] = -1;
		d[S] = 0, que.push ( S );
		while ( !que.empty () ) {
			int u = que.front (); que.pop ();
			for ( int i = head[u], v; i; i = graph[i].nxt ) {
				if ( graph[i].flow && !~d[v = graph[i].to] ) {
					d[v] = d[u] + 1;
					que.push ( v );
				}
			}
		}
		return ~d[T];
	}

	inline int dfs ( const int u, const int iflow ) {
		if ( u == T ) return iflow;
		int ret = 0;
		for ( int& i = curh[u], v; i; i = graph[i].nxt ) {
			if ( graph[i].flow && d[v = graph[i].to] == d[u] + 1 ) {
				int oflow = dfs ( v, imin ( iflow - ret, graph[i].flow ) );
				ret += oflow, graph[i].flow -= oflow, graph[i ^ 1].flow += oflow;
				if ( ret == iflow ) break;
			}
		}
		if ( !ret ) d[u] = -1;
		return ret;
	}

	inline int calc ( const int s, const int t ) {
		S = s, T = t;
		int ret = 0;
		for ( ; bfs (); ret += dfs ( S, INF ) ) {
			for ( int i = 0; i <= bound; ++i ) curh[i] = head[i];
		}
		return ret;
	}
} graph;

inline bool check ( const int lim ) {
	int cnt = n * m * 2;
	graph.bound = cnt + n + m + 3;
	graph.clear (), graph.S = cnt + n + m + 2, graph.T = graph.S + 1;
	int rS = 0, rT = cnt + n + m + 1;
	for ( int i = 0; i <= graph.bound; ++i ) deg[i] = 0;
	for ( int i = 1; i <= n; ++i ) { // row.
		int lw = imax ( 0, rsum[i] - lim ), up = rsum[i] + lim;
		deg[rS] -= lw, deg[cnt + i] += lw;
		graph ( rS, cnt + i, up - lw );
	}
	for ( int i = 1; i <= m; ++i ) { // col.
		int lw = imax ( 0, csum[i] - lim ), up = csum[i] + lim;
		deg[cnt + n + i] -= lw, deg[rT] += lw;
		graph ( cnt + n + i, rT, up - lw );
	}
	for ( int i = 1; i <= n; ++i ) {
		for ( int j = 1; j <= m; ++j ) {
			int id = ( i - 1 ) * m + j, rid = id + n * m;
			deg[id] -= L, deg[rid] += L;
			graph ( id, rid, R - L );
			graph ( cnt + i, id, INF ), graph ( rid, cnt + n + j, INF );
		}
	}
	int req = 0;
	for ( int i = rS; i <= rT; ++i ) {
		if ( deg[i] > 0 ) graph ( graph.S, i, deg[i] );
		else if ( deg[i] ) req -= deg[i], graph ( i, graph.T, -deg[i] );
	}
	graph ( rT, rS, INF );
	return graph.calc ( graph.S, graph.T ) == req;
}

int main () {
	scanf ( "%d %d", &n, &m );
	for ( int i = 1; i <= n; ++i ) {
		for ( int j = 1, a; j <= m; ++j ) {
			scanf ( "%d", &a );
			rsum[i] += a, csum[j] += a;
		}
	}
	scanf ( "%d %d", &L, &R );
	int l = 0, r = imax ( n, m ) * R;
	while ( l < r ) {
		int mid = l + r >> 1;
		if ( check ( mid ) ) r = mid;
		else l = mid + 1;
	}
	printf ( "%d\n", l );
	return 0;
}