题意
Sol
自己都不知道自己怎么做出来的系列
不难观察出几个性质：
- 最优策略一定是先把某一行弄黑，然后再用这一行去覆盖不是全黑的列
- 无解当且仅当无黑色。否则第一个黑色所在的行\(i\)可以先把第\(i\)列弄出一个黑色，接下来第\(i\)列的黑色可以把第\(i\)行全部弄成黑色。
然后直接算出把每一行弄黑的步数取个min就行了。
代码里面有注释。
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1001, INF = 1e9 + 1, mod = 1e9 + 7;
const double eps = 1e-9, pi = acos(-1);
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N;
char s[MAXN][MAXN];
int FullRow[MAXN], HaveRow[MAXN], FullCol[MAXN], HaveCol[MAXN];//full:每行每列是否全为1 / have:是否至少有一个1
signed main() {
N = read();
for(int i = 1; i <= N; i++) scanf("%s", s[i] + 1);
bool flag = 0;
fill(FullRow + 1, FullRow + N + 1, 1); fill(FullCol + 1, FullCol + N + 1, 1);
for(int i = 1; i <= N; i++)
for(int j = 1; j <= N; j++)
if(s[i][j] == '#') flag = 1, HaveRow[i] = 1, HaveCol[j] = 1;
else FullRow[i] = 0, FullCol[j] = 0;
if(!flag) return puts("-1"), 0;
int ans = INF;
for(int i = 1; i <= N; i++) {
int now = 0;
if(FullRow[i]) {
for(int j = 1; j <= N; j++) now += (!FullCol[j]);//如果第i行全为黑，答案就加上不为黑的列数
chmin(ans, now);
continue;
}
if(!HaveCol[i]) {
if(!HaveRow[i]) continue;
else now++;//用该行的一个黑去染黑对应的列
}
for(int j = 1; j <= N; j++) {
if(FullCol[j]) continue;
if(s[i][j] == '.') now += 2;
else now++;
}
chmin(ans, now);
}
cout << ans;
return 0;
}