odeforces 615A. Bulbs

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本文详细解析了 CodeForces 平台上的 615A 题目——Bulbs 的解题思路及实现代码。该题目要求通过一系列按钮操作使所有灯泡亮起，介绍了如何利用 C 语言来标记和检查灯泡状态，并最终判断是否能成功点亮所有灯泡。

codeforces 615A. Bulbs

A. Bulbs
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it’s connected to. Can Vasya light up all the bulbs?

If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input

The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.

Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output

If it’s possible to turn on all m bulbs print “YES”, otherwise print “NO”.
Examples
Input

3 4
2 1 4
3 1 3 1
1 2

Output

YES

Input

3 3
1 1
1 2
1 1

Output

Note

In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
题目大意：n个开关控制m个灯泡，n个开关能否把所有灯泡打开。

#include<stdio.h>
#include<stdbool.h>
#include<string.h>
bool vis[110];
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        int i,j;
        int x;
        int x1;
        ///将有开关控制的电灯标记为1
        for(i=0;i<n;i++)
        {
            scanf("%d",&x);
            while(x--)
            {
                scanf("%d",&x1);
                vis[x1]=1;
            }

        }
        ///判断是否所有灯泡被标记
        bool abc=1;
        for(j=1;j<=m;j++)
        {
            if(!vis[j])
            {
                abc=0;
                break;
            }

        }
        printf(abc? "YES\n":"NO\n");
    }
    return 0;
}