每个海面要么放要么不放，因此可以用二分图匹配， 考虑把同一行内的能互相看到的点放到一个行块里，同一列内能看到的点放到一个列块里，然后每一个行块都可以和该行块里所有海面的列块连边，选了这个行块，就必须选且只选择一个该行块里的一个海面对应的列块。
#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define N 10100
using namespace std;
struct edg {
int to, nex;
}e[N];
char ma[101][101];
int n, m, T, n1, n2, cnt, cntx, cnty;
int flag[N], lin[N], x[101][101], y[101][101], vis[N];
void add(int f, int t)
{
e[++cnt].to = t;
e[cnt].nex = lin[f];
lin[f] = cnt;
}
bool dfs(int now)
{
for (int i = lin[now]; i; i = e[i].nex)
{
int to = e[i].to;
if (vis[to]) continue;
vis[to] = 1;
if (flag[to] == -1 || dfs(flag[to]))
{
flag[to] = now;
return 1;
}
}
return 0;
}
void clea()
{
memset(flag, -1, sizeof(flag));
memset(e, 0, sizeof(e));
memset(lin, 0, sizeof(lin));
memset(vis, 0, sizeof(vis));
cnt = 0;
cntx = cnty = 1;
}
int main()
{
scanf("%d", &T);
while (T--)
{
clea();
scanf("%d%d", &n, &m);
char ca = getchar();;
for (int i = 1; i <= n; i++, ca = getchar())
for (int j = 1; j <= m; j++)
scanf("%c", &ma[i][j]);
for (int i = 1; i <= n; i++, cntx++)
for (int j = 1; j <= m; j++)
{
if (ma[i][j] == '*')//把属于同一行的数并到一起
x[i][j] = cntx;
if (ma[i][j] == '#')
++cntx;
}
for (int i = 1; i <= m; i++, cnty++)
for (int j = 1; j <= n; j++)
{
if (ma[j][i] == '*')
y[j][i] = cnty;
if (ma[j][i] == '#')
++cnty;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (ma[i][j] == '*')
add(x[i][j], y[i][j]);
n1 = cntx, n2 = cnty;
int ans = 0;
for (int i = 1; i <= n1; i++)
{
memset(vis, 0, sizeof(vis));
if (dfs(i))
ans++;
}
for (int i = 1; i <= n2; i++)
if (flag[i]!=-1)
printf("%d %d
", flag[i], i);
printf("%d
", ans);
}
return 0;
}//
/*
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*/