hdu5823 color II（状态压缩DP）

题意：给你一个图G，求每个子图的最少染色数

思路：每个子图的染色问题，可以看成是找每个子图的独立集问题（两两顶点互不相邻的图叫独立集），可以先枚举每个状态，然后枚举每个顶点，表示以这个顶点为起始的子图，再枚举每个顶点，表示这个顶点和上面那个顶点所构成的图是一个独立集，预处理出所有非法的状态，即他们之间有边。

           令dp[s]为子图S已经染色的最少染色数，那么有dp[s]=min(dp[s],dp[s^s0]+1)其中s0是S的子集，并且内部没有变，也就是s0是一个可以染成同一种颜色的结点集

#include<bits/stdc++.h>using namespace std;#define LL long long#define inf 0x3f3f3fLL dp[1<<18];char mp[20][20];int vis[1<<18];int main(){    int T;scanf("%d",&T);while(T--){memset(vis,0,sizeof(vis));int n;scanf("%d",&n);for(int i= 1;i<=n;i++)scanf(" %s",mp[i]+1);for(int i = 1;i<(1<<n);i++){for(int j = 0;j<n;j++){if(i&(1<<j)){for(int k = 0;k<n;k++){if(i&(1<<k) && mp[j+1][k+1]=='1'){vis[i]=1;break;}}}if(vis[i])break;}}dp[0]=0;for(int s = 1;s<(1<<n);s++){            dp[s]=inf;for(int s0 = s;s0;s0=(s0-1)&s){if(!vis[s0])dp[s]=min(dp[s],dp[s^s0]+1);}}LL ans = 0;LL f = 1;for(int i = 1;i<(1<<n);i++){f*=233;            ans +=(dp[i]*f);}printf("%u\n",(int)ans);}}

Problem 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)=∑v∈S2v. Let the chromatic number of S be fid(S).

You need to output ∑1<=id(S)<=2n−1fid(S)×233id(S)mod232.

 

Sample Input

24011010101101001040111101011011010

 

Sample Output

10224233542538351020
Hint
For the first test case, ans[1..15]= {1, 1, 2, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3}
 

 

Author

学军中学