Codeforce957B(暴力模拟)

最新推荐文章于 2025-06-26 23:56:53 发布

原创最新推荐文章于 2025-06-26 23:56:53 发布 · 464 阅读

0 ·

CC 4.0 BY-SA版权

题目笔记专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种算法，用于验证一系列操作是否能将初始状态的矩阵中的特定符号消除。通过使用标记辅助的方法，确保每次操作选择的行和列满足特定条件，并最终判断矩阵是否能被完全清除。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Mystical Mosaic

There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There’s another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that or , where denotes intersection of sets, and denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.tp://codeforces.com/contest/957/problem/B)

题意:给你一个n*m矩阵，有一种操作—在矩阵中选出一些行与列(每一行，每一列只能选一次)，要求这些行与列相交的点为‘#’，并且操作之后‘#’变为‘.’，可以进行多次操作，问是否能在多次操作后使得矩阵没有‘#’。

思路:直接暴力模拟操作，对每行每列标记，判断是否符合条件。

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
#include <functional>
#define inf 1000000000
using namespace std;
typedef long long ll;
const int MAXN=1e9+10;
const int MAX=1e3+10;
const double eps=1e-6;

int n,m;
char G[100][100];
int r[100],c[100];

int main(){
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    #endif

    cin>>n>>m;
    for(int i=0;i<n;i++)
        scanf("%s",G[i]);
    int juge=1;
    int cnt=1;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(G[i][j]=='#'){
                if(!r[i]&&!c[j]){
                    r[i]=c[j]=cnt++;
                }
                else if(!r[i]||!c[j]){
                    r[i]=c[j]=max(r[i],c[j]);
                }
                else if(r[i]&&c[j]&&r[i]!=c[j])
                    juge=0;
            }
        }
        if(!juge)   break;
    }
    for(int i =0;i<n;i++){
        for(int j=0;j<m;j++){
            if(G[i][j]=='.'&&r[i]==c[j]&&r[i]!=0){
                //cout<<i<<" "<<j<<endl;
                juge=0;
                break;
            }
        }
    }

    if(juge)    cout<<"Yes"<<endl;
    else cout<<"No"<<endl;


    return 0;
}