loj#6030. 「雅礼集训 2017 Day1」矩阵(贪心 构造)

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：

1. 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
2. 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;
}