Codeforces.985B.Switches and Lamps（思维）

最新推荐文章于 2022-06-21 17:57:19 发布

MiaoPlus

最新推荐文章于 2022-06-21 17:57:19 发布

阅读量380

点赞数 1

CC 4.0 BY-SA版权

分类专栏： ===========数学知识=========== 文章标签： Codeforces.985B Switches and Lamps

本文链接：https://blog.youkuaiyun.com/XxxxxM1/article/details/80470072

===========数学知识=========== 专栏收录该内容

40 篇文章

订阅专栏

本文探讨了在给定开关与灯泡连接矩阵的情况下，如何判断是否存在某个开关可以在被忽略时，通过操作其余所有开关使所有灯泡亮起的问题。通过遍历每个开关并检查其与灯泡的关系来寻找解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You are given n switches and m lamps. The i-th switch turns on some subset of the lamps. This information is given as the matrix aconsisting of n rows and m columns where ai, j = 1 if the i-th switch turns on the j-th lamp and ai, j = 0 if the i-th switch is not connected to the j-th lamp.

Initially all m lamps are turned off.

Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.

It is guaranteed that if you push all n switches then all m lamps will be turned on.

Your think that you have too many switches and you would like to ignore one of them.

Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other n - 1 switches then all the m lamps will be turned on.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of the switches and the number of the lamps.

The following n lines contain m characters each. The character ai, j is equal to '1' if the i-th switch turns on the j-th lamp and '0' otherwise.

It is guaranteed that if you press all n switches all m lamps will be turned on.

Output

Print "YES" if there is a switch that if you will ignore it and press all the other n - 1 switches then all m lamps will be turned on. Print "NO" if there is no such switch.

Examples
inputCopy
4 5
10101
01000
00111
10000
outputCopy
YES
inputCopy
4 5
10100
01000
00110
00101
outputCopy
NO

考点：暴力求解

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    char c;
    while(cin>>n>>m){
        int a[2007][2007];
        int cnt[2010];
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++){
            cin>>c;
            a[i][j]=c-'0';
            cnt[j]+=a[i][j];
        }
        for(int i=0;i<n;i++){
            int flag=1;
            for(int j=0;j<m;j++){
                if(a[i][j]==cnt[j]) {flag=0;break;}
            }
            if(flag)  return 0 * puts("YES");
        }
        cout<<"NO"<<endl;
    }
    return 0;
}