HDOJ2806 位运算

最新推荐文章于 2025-04-28 11:29:15 发布

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最新推荐文章于 2025-04-28 11:29:15 发布

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文章标签：位运算

本文链接：https://blog.youkuaiyun.com/computer_user/article/details/78363671

Selecting Problems

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 861 Accepted Submission(s): 147

Problem Description

The college of computer & software is going to hold an ACM contest. N (0<N<=1000) students have registered for the contest, and the teacher LCY has thought up M (0<M<=15) problems numbered from 1 to M. LCY is so familiar with the students that he can forecast the result of every student for the M problems.
Now he will select as many problems as possible from the M problems. And he want at least K (0<=K<=M) students to solve all problems. Will you compute the maximum number of problems he can choose? Output "0" if there is no selection can satisfy the condition.

Input

First line of each case contains three integers: N, M and K. Then N lines follow, and each line has the format:
Name P N1 N2 … NP.
Name (the length won't exceed 20), P (number of problems he/she can solve, 0<=P<=M), then P integers (represent the problems, 1-based).

Output

Output the maximum number of problems.

Sample Input

  
   2 6 2
zl1 6 1 2 3 4 5 6
zl2 5 1 2 3 4 5
2 3 2
zl1 2 2 3
zl2 1 1

Sample Output

5
0

Source

HDU 2009-4 Programming Contest

Recommend

lcy | We have carefully selected several similar problems for you: 2807 2808 2801 2810 2809

分析：每个同学可以A的题构成一个集合，预处理该集合，通过位运算高效的枚举该集合所有的子集。

同样的，所有题目构成一个集合，也是位运算枚举集合的每个子集。

令s为一个集合，枚举方法如下

for(int ss=s; ss; ss=(ss-1)&s) {...}

代码如下：

#include <cstdio>
#include <cstring>
using namespace std;
int num[1<<16];
int dp[1<<16];
int cnt[1<<16];
int n,m,k;

void init(){
   char s[25];
   int p;
   int ans,x;
   memset(dp,0,sizeof(dp));
   memset(cnt,0,sizeof(cnt));

   for (int i=1; i<=n; i++) {
       scanf("%s",s);
       scanf("%d",&p);
       ans = 0;
       for (int j=1; j<=p ;j++) {
           scanf("%d",&x);
           ans |= 1<<(x-1);
       }
       cnt[ans]++;
   }
}

inline int MAX(int x, int y){return x>y?x:y;}

void solve(){
    int ans = 0;
    for (int i=1<<m; i>=0; i--) if (cnt[i]) {
        for (int j=i; j; j=(j-1)&i) dp[j]+=cnt[i];
    }

    for (int i=0; i<1<<m; i++) if (dp[i]>=k) ans = MAX(ans,num[i]);
    printf("%d\n",ans);
}

int main(){
    memset(num,0,sizeof(num));
    for (int i=0; i<1<<16; i++) {
        for (int j=0; j<16; j++) if (i&(1<<j)) num[i]++;
    }

    while (scanf("%d %d %d",&n,&m,&k)!=EOF){
        init();
        solve();
    }
    return 0;
}