AcWing91 最短Hamilton路径 (状压dp)

Posted 2020-12-20 tuchen

https://www.acwing.com/problem/content/93/

(dp[S][i]) 表示经过点的状态为 (S) , 当前在 (i) 点时的最短路
时间复杂度(O(n^2*2^n))

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 21;

int n;
int dp[1<<maxn][maxn], a[maxn][maxn];

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<‘0‘ || ch>‘9‘){ if(ch==‘-‘) f=-1; ch=getchar(); } while(ch>=‘0‘ && ch<=‘9‘){ s=s*10+ch-‘0‘; ch=getchar(); } return s*f; }

int main(){
	n = read();
	for(int i=0;i<n;++i) for(int j=0;j<n;++j) a[i][j] = read(); 
	
	memset(dp,0x3f,sizeof(dp));
	dp[1][0] = 0;
	
	for(int S=1;S<(1<<n);++S){
		for(int j=0;j<n;++j){
			if((S>>j) & 1){
				for(int k=0;k<n;++k){
					if(((S ^ (1<<j)) >> k) & 1){
						dp[S][j] = min(dp[S][j], dp[S^(1<<j)][k] + a[k][j]); 
					}
				}
			}
		}
	}
	
	printf("%d
",dp[(1<<n)-1][n-1]);
	
	return 0;
}