HDU5823 color II

 

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Description

You are given an undirected graph with n vertices numbered 0 through n-1. 

Obviously, the vertices have 2^n - 1 non-empty subsets. For a non-empty subset S, we define a proper coloring of S is a way to assign each vertex in S a color, so that no two vertices in S with the same color are directly connected by an edge. Assume we've used k different kinds of colors in a proper coloring. We define the chromatic number of subset S is the minimum possible k among all the proper colorings of S. 

Now your task is to compute the chromatic number of every non-empty subset of the n vertices.

Input

First line contains an integer t. Then t testcases follow. 

In each testcase: First line contains an integer n. Next n lines each contains a string consisting of '0' and '1'. For 0<=i<=n-1 and 0<=j<=n-1, if the j-th character of the i-th line is '1', then vertices i and j are directly connected by an edge, otherwise they are not directly connected. 

The i-th character of the i-th line is always '0'. The i-th character of the j-th line is always the same as the j-th character of the i-th line. 

For all testcases, 1<=n<=18. There are no more than 100 testcases with 1<=n<=10, no more than 3 testcases with 11<=n<=15, and no more than 2 testcases with 16<=n<=18.

Output

For each testcase, only print an integer as your answer in a line. 

This integer is determined as follows: 
We define the identity number of a subset S is id(S)=\sum_{v\in S} 2^v. Let the chromatic number of S be f_{id(S)}. 

You need to output \sum_{1<=id(S)<=2^n - 1} f_{id(S)} \times 233^{id(S)} \mod 2^{32}.

Sample Input

2
4
0110
1010
1101
0010
4
0111
1010
1101
1010

Sample Output

1022423354
2538351020


        
 

Hint

 

For the first test case, ans[1..15]= {1, 1, 2, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3}

Source

2016 Multi-University Training Contest 8

 

统计每个子集的最小染色数，按公式算就行了。

模拟染色是行不通的，因为在同一个图中根据不同的染色顺序，可能得出不同的答案。

正解是状压DP。

先预处理出每个子集是不是独立集，然后尽量把独立集中的点都染上同一个颜色，这样的染色方法是最优的。

然后在不同的独立集之间状态转移即可。

位运算大法好，位运算大法好。

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
using namespace std;
const LL mod=4294967296;
const int mxn=320000;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int f[mxn];
int n;
LL ans;
int mp[20];
bool inv[mxn];
int main(){
    int T;
    T=read();
    int i,j;
    while(T--){
        memset(mp,0,sizeof mp);
        memset(f,0,sizeof f);
        memset(inv,0,sizeof inv);
        ans=0;
        n=read();
        char ch[30];
        int ed=(1<<n);
        for(i=0;i<n;++i){
            scanf("%s",ch);
            for(j=0;j<n;++j){
                if(ch[j]=='1')mp[i]|=(1<<j);
            }
        }
        
        for(int s=1;s<ed;++s){
            bool flag=1;
            for(i=0;i<n;++i){
                if(s&(1<<i)){
                    if(s & mp[i]){
                        flag=0;
                        break;
                    }
                }
            }
            inv[s]=flag;
        }
        for(int s=1;s<ed;++s){
            int tmp=1e6;
            for(i=s;i;i=(i-1)&s){
                if(inv[i]){
                    tmp=min(tmp,f[s^i]+1);
                }
            }
            f[s]=tmp;
        }
        LL tmp=1;
        for(i=1;i<ed;i++){
            tmp=(tmp*233)%mod;
            ans=(ans+tmp*f[i])%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}