LOJ Chen 6030. "Yali training 2017 Day1" matrix (greedy construction)

meaning of the title

link

Sol

I don't know how to make the series

It is not difficult to observe several properties:

  1. The best strategy must be to black a row first, and then use this row to cover columns that are not all black
  2. No solution if and only if there is no black. Otherwise, the row \ (i \) where the first black is located can first make a black in column \ (i \), and then the black in column \ (i \) can make all the rows \ (i \) black.

Then calculate the steps of blackening each line directly and take a min ute.

There are comments in the code.

#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: whether each row and column is all 1 / have: whether there is at least one 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]);//If line i is all black, add the number of columns that are not black to the answer
            chmin(ans, now); 
            continue;
        }
        if(!HaveCol[i]) {
            if(!HaveRow[i]) continue;
            else now++;//Use a black of the row to color the corresponding column
        }
        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;
}

Tags: C++

Posted on Tue, 03 Dec 2019 14:02:11 -0500 by barryflood22