【UVA11825】Hackers' Crackdown

题意

  Miracle Corporations has a number of system services running in a distributed computer system which is a prime target for hackers. The system is basically a set of N computer nodes with each of them running a set of N services. Note that, the set of services running on every node is same everywhere in the network. A hacker can destroy a service by running a specialized exploit for that service in all the nodes.
  One day, a smart hacker collects necessary exploits for all these N services and launches an attack on the system. He finds a security hole that gives him just enough time to run a single exploit in each computer. These exploits have the characteristic that, its successfully infects the computer where it was originally run and all the neighbor computers of that node.
  Given a network description, find the maximum number of services that the hacker can damage.

题面

  有n个节点，每一个节点都有n种标记，现在可以对每一个节点分别进行一次操作，每一次操作可以选择一种标记，使这个节点及与其相邻的节点覆盖这种标记（一个节点可以被覆盖多种标记），问经过n操作后，最多有多少种标记覆盖了所有节点
  n≤16，节点编号从0开始

解法

子集DP：
  第一次做子集DP的题……
  这个题目的意思就是说：全集为U，|U|=n，有n个集合，为{P1，P2……Pn}，现在要对这n个集合进行分组，要求P1∪P2∪……∪Pk=U，P为该组内的集合，求最多能分多少组
  设stai为一个二进制数，表示第i个节点相邻的点及自身，例如i=10_（2），stai=10110_（2），表示与1号节点相邻的有4，2两个节点。由此可知，stai表示对i进行操作能够标记的点的集合
  设pi表示选择i表示的集合能够标记的点集，i为一个二进制数，例如i=10010_（2）表示选择{0，4}这个集合能够标记的点集

  设fi表示选择i集合能够标记的全集个数，即答案，i也是一个二进制数。状态已经设出来了，下面就是如何转移：
  对于一种状态i，i表示的就是一个集合，考虑如何枚举出i的所有子集，可以参见刘汝佳的紫书

复杂度

O（2n*平均非空子集个数），当有n个物品时，总非空子集数为：∑ni=1Cin

代码

#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#define Lint long long intusing namespace std;const int N=20;const int L=(1<<17);int f[L+5],p[L+5];int sta[N];int n,A;int main(){    int len,x,C=0;    while( scanf("%d",&n) && n )    {        memset( sta,0x0,sizeof sta );        memset( f,0x0,sizeof f ),memset( p,0x0,sizeof p );        for(int i=0;i<=n-1;i++)        {            scanf("%d",&len);            for(int j=1;j<=len;j++)            {                scanf("%d",&x);                sta[i]|=(1<<x);            }            sta[i]|=(1<<i);        }        A=(1<<n)-1;        for(int i=0;i<=A;i++)        {            p[i]=0;            for(int j=0;j<=n-1;j++)                if( i&(1<<j) )                    p[i]|=sta[j];        }        for(int i=0;i<=A;i++)        {            if( p[i]!=A )   f[i]=0;            else   f[i]=1;            for(int j=i; j ;j=(j-1)&i)                f[i]=max( f[i],f[j]+f[i^j] );        }        printf("Case %d: %d\n",++C,f[A]);    }    return 0;}