Winner Winner
题目描述
The FZU Code Carnival is a programming competetion hosted by the ACM-ICPC Training Center of Fuzhou University. The activity mainly includes the programming contest like ACM-ICPC and strive to provide participants with interesting code challenges in the future.
Before the competition begins, YellowStar wants to know which teams are likely to be winners. YellowStar counted the skills of each team, including data structure, dynamic programming, graph theory, etc. In order to simplify the forecasting model, YellowStar only lists M skills and the skills mastered by each team are represented by a 01 sequence of length M. 1 means that the team has mastered this skill, and 0 does not.
If a team is weaker than other teams, this team cannot be a winner. Otherwise, YellowStar thinks the team may win. Team A(a1, a2, ..., aM ) is weaker than team B(b1, b2, ..., bM ) if ∀i ∈ [1, M], ai ≤ bi and ∃i ∈ [1, M], ai < bi.
Since YellowStar is busy preparing for the FZU Code Carnival recently, he dosen’t have time to forecast which team will be the winner in the N teams. So he asks you to write a program to calculate the number of teams that might be winners.
输入
Input is given from Standard Input in the following format:
N M
s1 s2 . . . sN
The binary representation of si indicates the skills mastered by teami.
Constraints
1 ≤ N ≤ 2 × 106
1 ≤ M ≤ 20
0 ≤ si < 2M
输出
Print one line denotes the answer.
样例输入
3 3
2 5 6
样例输出
2
题意：
有n个队伍从1~n 有m种技能 十进制数转二进制表示每个队伍对技能的熟练程度 如果是1表示完全掌握 如果是0 没完全掌握 如果任意两支队伍 其中一个的每种技能都<=另外一支队伍 那么这个队伍不可能获胜 输出最终可能获胜的队伍数目
例如： 2 5 6
二进制表示为： 0 1 0
1 0 1
1 1 0
每一列代表一支队伍 分别为1、2、3
通过比较1 3 可以把3排除 其余的无论怎么比较都无法再排除 所以最终结果就是 2支队伍 分别是： 1、 2
思路：
看了题解说是dp 感觉也不是dp 就是先记录每个队伍 然后遍历排除其他的队伍 最后统计还剩多少不能排除的
例如： n=3 m=3 2 5 6
从最大值开始循环 6的二进制 1 1 0 可以看出来 6可以覆盖1 0 0 和 0 1 0 和 0 0 0 三个数分别为4 2 0 把它们的vis值变为-1 最后只统计vis值为正的 即是结果
AC代码：
#include<bits/stdc++.h>
using namespace std;
int vis[(1<<(20))+5];
int main()
{
int n,m,num;
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++){
scanf("%d",&num);
vis[num]++;
}
int ans=0;
for(int i=(1<<m)-1;i>=0;i--){
if(vis[i]!=0){
if(vis[i]>0){
ans+=vis[i]; ///vis[]为正ans直接加上
}
for(int j=m-1;j>=0;j--){
if(i&(1<<j)){
vis[i^(1<<j)]=-1; ///我感觉该是 vis[(1<<j)]赋值为 -1 但不知道为什么不可
} /// 以这样
}
}
}
printf("%d
",ans);
return 0;
}