luogu P4194 矩阵

嘟嘟嘟




先二分。

令二分的值为\(mid\)，则对于每一行都要满足\(|\sum_{i = 1} ^ {n} (A_{ij} - B_{ij})|\)，把绝对值去掉，就得到了\((\sum_{i = 1} ^ {n} A_{ij}) - mid \leqslant \sum_{i = 1} ^ {n} B_{ij} \leqslant (\sum_{i = 1} ^ {n} A_{ij}) + mid\)。（列同理）

这就很明显了，因为是网格图，所以每一行每一列看成一个点建立二分图，从源点向每一行连容量为\([(\sum_{i = 1} ^ {n} A_{ij}) - mid， (\sum_{i = 1} ^ {n} A_{ij}) + mid]\)的边，然后每一列向汇点也这么连边。同时每一行向每一列连容量为\([L,R]\)的边。

然后跑上下界网络流。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 205;
const int maxe = 1e6 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, l, r, s, t, S, T;
int a[maxn][maxn], sumn[maxn], summ[maxn];
struct Edge
{
  int nxt, to, cap, flow;
}e[maxe];
int head[maxn << 1], ecnt = -1;
In void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w, 0};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], x, 0, 0};
  head[y] = ecnt;
}

int dis[maxn << 1];
In bool bfs()
{
  Mem(dis, 0); dis[S] = 1;
  queue<int> q; q.push(S);
  while(!q.empty())
    {
      int now = q.front(); q.pop();
      for(int i = head[now], v; ~i; i = e[i].nxt)
	{
	  if(!dis[v = e[i].to] && e[i].cap > e[i].flow)
	    dis[v] = dis[now] + 1, q.push(v);
	}
    }
  return dis[T];
}
int cur[maxn << 1];
In int dfs(int now, int res)
{
  if(now == T || res == 0) return res;
  int flow = 0, f;
  for(int& i = cur[now], v; ~i; i = e[i].nxt)
    {
      if(dis[v = e[i].to] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
	{
	  e[i].flow += f; e[i ^ 1].flow -= f;
	  flow += f; res -= f;
	  if(res == 0) break;
	}
    }
  return flow;
}

In int maxflow()
{
  int flow = 0;
  while(bfs())
    {
      memcpy(cur, head, sizeof(head));
      flow += dfs(S, INF);
    }
  return flow;
}

int d[maxn << 1], tot = 0;
In void build(int lim)
{
  Mem(head, -1); ecnt = -1; Mem(d, 0); tot = 0;
  for(int i = 1; i <= n; ++i)
    {
      int tp = max(sumn[i] - lim, 0);
      d[s] += tp, d[i] -= tp;
      addEdge(s, i, sumn[i] + lim - tp);
    }
  for(int i = 1; i <= m; ++i)
    {
      int tp = max(summ[i] - lim, 0);
      d[t] -= tp, d[i + n] += tp;
      addEdge(i + n, t, summ[i] + lim - tp);
    }
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      {
	d[i] += l, d[j + n] -= l;
	addEdge(i, j + n, r - l);
      }
  for(int i = 0; i <= t; ++i)
    if(d[i] >= 0) addEdge(i, T, d[i]), tot += d[i];
    else addEdge(S, i, -d[i]);
  addEdge(t, s, INF);
}

In bool judge(int x)
{
  build(x);
  return maxflow() == tot;
}

int main()
{
  Mem(head, -1);
  n = read(), m = read(); s = 0, t = n + m + 1;
  S = t + 1, T = t + 2;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) a[i][j] = read();
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) sumn[i] = sumn[i] + a[i][j];
  for(int j = 1; j <= m; ++j)
    for(int i = 1; i <= n; ++i) summ[j] = summ[j] + a[i][j];
  l = read(), r = read();
  int L = 0, R = 1e8;
  while(L < R)
    {
      int mid = (L + R) >> 1;
      if(judge(mid)) R = mid;
      else L = mid + 1;
    }
  write(L), enter;
  return 0;
}