hdu 3006 状压

最新推荐文章于 2020-05-17 10:44:24 发布

nielaishan

最新推荐文章于 2020-05-17 10:44:24 发布

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本文介绍了一个基于二进制表示的算法问题，通过使用状态压缩（状压）技巧来解决如何计算由给定多个集合合并产生的不同新集合的数量。文章提供了一段C++代码示例，演示了如何利用二进制位运算实现这一过程。

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The Number of set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1417 Accepted Submission(s): 883

Problem Description
Given you n sets.All positive integers in sets are not less than 1 and not greater than m.If use these sets to combinate the new set,how many different new set you can get.The given sets can not be broken.

Input
There are several cases.For each case,the first line contains two positive integer n and m(1<=n<=100,1<=m<=14).Then the following n lines describe the n sets.These lines each contains k+1 positive integer,the first which is k,then k integers are given. The input is end by EOF.

Output
For each case,the output contain only one integer,the number of the different sets you get.

Sample Input
4 4
1 1
1 2
1 3
1 4
2 4
3 1 2 3
4 1 2 3 4

Sample Output
15
2

题意

给出的集合，经过合并，能生成多少个集合，包含原来的集合。

解析

结合二进制表示，数k表示为第k位为1，集合{1, 2, 4}可以表示为1011，若有集合{3, 4},表示1100，这两个集合合并后为{1, 2, 3, 4},二进制表示为1111，即为1011 | 1100 = 1111

代码

//
//  main.cpp
//  0000
//  状压，二进制表示mi是否存在
//
//  Created by nielaishan on 2017/2/23.
//  Copyright © 2017年 用户. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int bit[1<<14];
int main(int argc, const char * argv[]) {
    int n, m;
    while (scanf("%d%d", &n, &m)!=EOF) {
        int w;
        memset(bit, 0, sizeof(bit));

        for (int i=0; i<n; i++) {
            int k;
            scanf("%d", &k);
            w = 0;
            for (int j=0; j<k; j++) {
                int a;
                scanf("%d", &a);
                w += 1<<(a-1);
            }
            bit[w] = 1;
            for (int j=0; j<1<<14; j++) {
                if (bit[j])
                    bit[j|w] = 1;
            }
        }
        int ans = 0;
        for (int i=0; i<1<<14; i++)
            if (bit[i])
                ans++;
        printf("%d\n", ans);
    }
    return 0;
}