Garlic Jun's Crayons

Garlic Jun received a birthday present - a box of beautiful crayons, which made him very happy. After completing a graph theory topic, Mr. Garantou took a crayon and wanted to color an undirected graph on the topic. There are a total of n on the undirected graph $n$  points, numbered from $0$  to $n-1$ , then the graph will have 2^n - 12n −1 non-empty subgraphs.

Mr. Garlic wants to give every  $Color\; the\; i$  subgraphs, so that the color of the two points connected by any edge is different, and now he wants to know how to give the $Coloring\; of\; i$  sub-images, how many different colors are required at least, denoted as s_isi​. Mr. Garantou would like to ask you to help him calculate the result of the following formula:

\displaystyle \sum_{i = 1}^{2^n-1} s_i \times 233^{i}.i=1∑2n−1​si​×233i.

input format

Enter an integer  $n$（$1\le n\le 16$ ), indicating that the undirected graph has $n$  points.

Next enter a  $n\times n$ 01 _ $01$  matrix, representing the case of undirected graph edges. if $a\left[i\right]\left[j\right]=1$ , then $i$  give jThere are edges between $j$$a\left[i\right]\left[j\right]=0$ , it means $i$ give j There are no edges connected between$j\; .$

output format

Output a line, output the result of the above formula, the result may be very large, the output result is  2^{32}23 2  take the remainder of the result.

sample input

4
0111
1011
1101
1110

Sample output

1595912448

Problem-solving instructions: Initialize all subgraphs that are painted with only one color. Remember such a subgraph dp[i]=1; the subgraph with fewer nodes is to update the subgraph with all nodes,

dp[i]=min(dp[i],dp[j^i]+dp[j]); Add the coloring numbers of the two sub-atlases of the graph. For example, a picture such as 1-2-3-4-5-6, we will get dp[picture: 1-3-5]=1, dp[picture: 2-4-6]=1 during preprocessing

Then dp[1-2-3-4-5-6] will become 2 when updated;

Nothing to say, this question stuck me for a long time! The local is right, the online submission of the error indicates that there is a problem with oj! . . . . . I forgot the initial. Every round of i cycle mn needs to be set to infinity, heck.

AC code:

#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstdio>
using namespace std;
const int inf=0x3f3f3f3f;
int dp[1<<16];
int a[20][20];
bool mark[1<<16];
unsigned int powdiy(int x,int y){
	unsigned int re=1,be=x;
	while(y){
		if(y&1)re*=be;
		be*=be;
		y>>=1;
	}
	return re;
}
void judge(int x){
	int tot = 1;
	int cot=0;
	int note[20];
	int temp=x;
	while(x){
		if(x&1){
			note[cot++]=tot;
		}
		x>>=1;
		to++;
	}
	for(int i=0;i<cot-1;i++)
	 for(int j=i+1;j<cot;j++){
	 	if(a[note[i]][note[j]]){
	 		return ;
		 }
	 }
	 dp[temp]=1;
	 return ;
}
unsigned int  sum=0;
int main(){
	int n;cin>>n;
	memset(dp,inf,sizeof(dp));
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
		   scanf("%1d",&a[i][j]);
	}
	for(int i=1;i<(1<<n);i++){
		judge(i);
	}
	for(int i=1;i<(1<<n);i++){
		int mn=inf;
		for(int j=i;j;j=(j-1)&i){
			dp[i]=min(dp[i],dp[j^i]+dp[j]);
			mn = min (dp [i], mn);
		}
		sum+=(mn*powdiy(233,i));
	}
	cout<<sum<<endl;
	return 0;
}