ZOJ 3471 Most Powerful（状压dp）_most powerful csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/44257207

本文深入探讨了AI算法如何与大数据开发相结合，通过实际案例展示了如何利用AI技术优化大数据处理流程，实现数据价值的最大化。主要内容包括AI在数据预处理、特征提取、模型训练与优化等阶段的应用，以及大数据技术在AI项目实施过程中的关键作用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Recently, researchers on Mars have discovered N powerful atoms. All of them are different. These atoms have some properties. When two of these atoms collide, one of them disappears and a lot of power is produced. Researchers know the way every two atoms perform when collided and the power every two atoms can produce.

You are to write a program to make it most powerful, which means that the sum of power produced during all the collides is maximal.

Input

There are multiple cases. The first line of each case has an integer N (2 <= N <= 10), which means there are N atoms: A₁, A₂, ... , A_N. Then N lines follow. There are N integers in each line. The j-th integer on the i-th line is the power produced when A_i and A_j collide with A_j gone. All integers are positive and not larger than 10000.

The last case is followed by a 0 in one line.

There will be no more than 500 cases including no more than 50 large cases that N is 10.

Output

Output the maximal power these N atoms can produce in a line for each case.

Sample Input

2
0 4
1 0
3
0 20 1
12 0 1
1 10 0
0

Sample Output

4
22

题意: n中气体，a[i][j]表示 i 和j撞后 j消失得到的价值，求最大得到的价值为多少？

思路：状压dp，从后往前推状态

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef long long ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 11

ll dp[1<<11];
ll a[N][N];
int n;

void solve()
{
	int i,j,cur;
	ll ans=0;
	int len=1<<n;

	for(cur=len-1;cur>=0;cur--)  //枚举当前的状态
		fre(i,1,n+1)                     //有 i 气体
	   {
		  if(!(cur&(1<<(i-1)))) continue;

		  fre(j,1,n+1)                   //有 j 气体
		    {
                if(i==j) continue;
		    	if(!(cur&(1<<(j-1)))) continue;

		        int to=cur^(1<<(j-1));    //a[i][j] j 消失后状态

                dp[to]=max(dp[to],dp[cur]+a[i][j]);

                if(ans<dp[to])
				 ans=dp[to];

		    }

	   }

     pf("%lld\n",ans);

}

int main()
{
	int i,j;
	while(sf(n),n)
	{
		fre(i,1,n+1)
		 fre(j,1,n+1)
		 {
		 	scanf("%lld",&a[i][j]);
		 }

        mem(dp,0);

        solve();
	}
   return 0;
}