| 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.
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.
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_{vin 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)} imes 233^{id(S)} mod 2^{32}.
This integer is determined as follows:
We define the identity number of a subset S is id(S)=sum_{vin 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)} imes 233^{id(S)} mod 2^{32}.
Sample Input
2
4
0110
1010
1101
0010
4
0111
1010
1101
1010
Sample Output
1022423354
2538351020
Hint
Source
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。
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #define LL long long 8 using namespace std; 9 const LL mod=4294967296; 10 const int mxn=320000; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 int f[mxn]; 18 int n; 19 LL ans; 20 int mp[20]; 21 bool inv[mxn]; 22 int main(){ 23 int T; 24 T=read(); 25 int i,j; 26 while(T--){ 27 memset(mp,0,sizeof mp); 28 memset(f,0,sizeof f); 29 memset(inv,0,sizeof inv); 30 ans=0; 31 n=read(); 32 char ch[30]; 33 int ed=(1<<n); 34 for(i=0;i<n;++i){ 35 scanf("%s",ch); 36 for(j=0;j<n;++j){ 37 if(ch[j]=='1')mp[i]|=(1<<j); 38 } 39 } 40 41 for(int s=1;s<ed;++s){ 42 bool flag=1; 43 for(i=0;i<n;++i){ 44 if(s&(1<<i)){ 45 if(s & mp[i]){ 46 flag=0; 47 break; 48 } 49 } 50 } 51 inv[s]=flag; 52 } 53 for(int s=1;s<ed;++s){ 54 int tmp=1e6; 55 for(i=s;i;i=(i-1)&s){ 56 if(inv[i]){ 57 tmp=min(tmp,f[s^i]+1); 58 } 59 } 60 f[s]=tmp; 61 } 62 LL tmp=1; 63 for(i=1;i<ed;i++){ 64 tmp=(tmp*233)%mod; 65 ans=(ans+tmp*f[i])%mod; 66 } 67 printf("%lld ",ans); 68 } 69 return 0; 70 }