#6030. 「雅礼集训 2017 Day1」矩阵 [贪心]

无解肯定是一个都没有。
首先你要想到一个贪心策略
就是要染满一列，然后再染其他列
很显然可以用 (k) 行染 (k) 列，然后你发现，如果第 (k) 列没有？
随便挑个来染上，多1的代价。
最后的话，看一下有多少个没染完的，搞下就完了

只能scanf，wdnmd

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {
    if (x < y)
        x = y;
}
Tp<class T> void cmin(T& x, const T& y) {
    if (x > y)
        x = y;
}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); }
Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN {
    char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
    char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
    FILEIN& operator>>(char& c) {
        while (isspace(c = GETC()))
            ;
        return *this;
    }
    FILEIN& operator>>(string& s) {
        while (isspace(ch = GETC()))
            ;
        s = ch;
        while (!isspace(ch = GETC())) s += ch;
        return *this;
    }
    Tp<class T> void read(T& x) {
        bool sign = 0;
        while ((ch = GETC()) < 48) sign ^= (ch == 45);
        x = (ch ^ 48);
        while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48);
        x = sign ? -x : x;
    }
    FILEIN& operator>>(int& x) { return read(x), *this; }
    FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {
    const static int LIMIT = 1 << 22;
    char quq[SZ], ST[233];
    int sz, O;
    ~FILEOUT() { flush(); }
    void flush() {
        fwrite(quq, 1, O, stdout);
        fflush(stdout);
        O = 0;
    }
    FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
    FILEOUT& operator<<(string str) {
        if (O > LIMIT)
            flush();
        for (char c : str) quq[O++] = c;
        return *this;
    }
    Tp<class T> void write(T x) {
        if (O > LIMIT)
            flush();
        if (x < 0) {
            quq[O++] = 45;
            x = -x;
        }
        do {
            ST[++sz] = x % 10 ^ 48;
            x /= 10;
        } while (x);
        while (sz) quq[O++] = ST[sz--];
    }
    FILEOUT& operator<<(int x) { return write(x), *this; }
    FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long

const int maxn = 1e3 + 31;
char a[maxn][maxn];
signed main() {
    // code begin.
    int n;
    scanf("%lld", &n);
    bool v = 0;
    vector<int> cntx(n + 3, 0), cnty(n + 3, 0);
    rep(i, 1, n) {
        scanf("%s", a[i] + 1);
        rep(j, 1, n) {
            if (a[i][j] == '#') {
                v = 1;
                ++cntx[i], ++cnty[j];
            }
        }
    }
    if (!v)
        out << -1 << '
';
    else {
        int ans = n;
        rep(i, 1, n) if (cnty[i]) cmin(ans, n - cntx[i]);
        else cmin(ans, n - cntx[i] + 1);
        rep(i, 1, n) if (cnty[i] != n)++ ans;
        out << ans << '
';
    }
    return 0;
    // code end.
}