题目链接：点击打开链接

思路：

我们首先假设这个图都是全0的

用n个点代表行，m个点代表列

用源点向行连一个值x 表示每行1的个数，向列连一个y表示每列y个1

则若行i和列j之间流过一个流量就表示 (i,j) 点填了1

那么若原来图中(i,j)点为0 则花费就是1

若原图中(i,j)点是1，则花费是-1

如此枚举x跑个费用流就好了

==居然把我多年的白书费用流坑掉了。。。

zkw走起啊

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;
#define maxn 110
#define maxm 100000
#define inf 0x3f3f3f3f
struct MaxFlow
{
	int size, n;
	int st, en, maxflow, mincost;
	bool vis[maxn];
	int net[maxn], pre[maxn], cur[maxn], dis[maxn];
	std::queue <int> Q;
	struct EDGE
	{
		int v, cap, cost, next;
		EDGE(){}
		EDGE(int a, int b, int c, int d)
		{
			v = a, cap = b, cost = c, next = d;
		}
	}E[maxm<<1];
	void init(int _n)
	{
		n = _n, size = 0;
		memset(net, -1, sizeof(net));
	}
	void add(int u, int v, int cap, int cost)
	{
		E[size] = EDGE(v, cap, cost, net[u]);
		net[u] = size++;
		E[size] = EDGE(u, 0, -cost, net[v]);
		net[v] = size++;
	}
	bool adjust()
	{
		int v, min = inf;
		for(int i = 0; i <= n; i++)
		{
			if(!vis[i])
				continue;
			for(int j = net[i]; v = E[j].v, j != -1; j = E[j].next)
				if(E[j].cap)
					if(!vis[v] && dis[v]-dis[i]+E[j].cost < min)
						min = dis[v] - dis[i] + E[j].cost;
		}
		if(min == inf)
			return false;
		for(int i = 0; i <= n; i++)
			if(vis[i])
				cur[i] = net[i], vis[i] = false, dis[i] += min;
		return true;
	}
	int augment(int i, int flow)
	{
		if(i == en)
		{
			mincost += dis[st] * flow;
			maxflow += flow;
			return flow;
		}
		vis[i] = true;
		for(int j = cur[i], v; v = E[j].v, j != -1; j = E[j].next)
		{
			if(!E[j].cap)
				continue;
			if(vis[v] || dis[v]+E[j].cost != dis[i])
				continue;
			int delta = augment(v, std::min(flow, E[j].cap));
			if(delta)
			{
				E[j].cap -= delta;
				E[j^1].cap += delta;
				cur[i] = j;
				return delta;
			}
		}
		return 0;
	}
	void spfa()
	{
		int u, v;
		for(int i = 0; i <= n; i++)
			vis[i] = false, dis[i] = inf;
		dis[st] = 0;
		Q.push(st);
		vis[st] = true;
		while(!Q.empty())
		{
			u = Q.front(), Q.pop();
			vis[u] = false;
			for(int i = net[u]; v = E[i].v, i != -1; i = E[i].next)
			{
				if(!E[i].cap || dis[v] <= dis[u] + E[i].cost)
					continue;
				dis[v] = dis[u] + E[i].cost;
				if(!vis[v])
				{
					vis[v] = true;
					Q.push(v);
				}
			}
		}
		for(int i = 0; i <= n; i++)
			dis[i] = dis[en] - dis[i];
	}
	int zkw(int s, int t, int need)
	{
		st = s, en = t;
		spfa();
		mincost = maxflow = 0;
		for(int i = 0; i <= n; i++)
			vis[i] = false, cur[i] = net[i];
		do
		{
			while(augment(st, inf))
				memset(vis, false, sizeof(vis));
		}while(adjust());
		if(maxflow < need)
			return -1;
		return mincost;
	}
}zkw;
char mp[50][50];
int n, m, siz;
int work(int x){
	int all = n*x;
	if(all % m || all > n*m)return inf;
	int y = all / m;
	zkw.init(n+m+2);
	int from = n+m+1, to = n + m + 2;
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
			if(mp[i][j]=='1')
				zkw.add(i, n+j, 1, -1);
			else 
				zkw.add(i, n+j, 1, 1);
	}
	for(int i = 1; i <= n; i++)
		zkw.add(from, i, x, 0);
	for(int i = 1; i <= m; i++)
		zkw.add(n+i, to, y, 0);
	return siz + zkw.zkw(from, to, -1);
}
void input(){	
	scanf("%d %d",&n,&m);
	siz = 0;
	for(int i = 1; i <= n; i++) {
		scanf("%s", mp[i]+1);
		for(int j = 1; j <= m; j++)
			siz += mp[i][j] == '1';
	}
}
int main(){
	int Cas = 0, T; scanf("%d",&T);
	while(T--){
		input();
		int ans = inf;
		for(int i = 0; i <= m; i++)
			ans = min(ans, work(i));
		printf("Case %d: %d\n", ++Cas, ans);
	}
	return 0;
}