# Install tensorflow Error: No Matching Distribution Found for TensorFlownire

2023-01-23   ES

Most Powerful

Time Limit: 2 Seconds
Memory Limit: 65536 KB

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, … , AN. 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 Ai and Aj collide with Aj 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

Author:
GAO, Yuan

Contest: ZOJ Monthly, February 2011

Title: N particles, AI and AJ collisions will generate AIJ energy at the same time AJ disappear, to the maximum energy that can be produced in the end

Question analysis: After seeing the amount of data, the obvious pressure is obvious. DP [i] indicates the maximum energy when the status is i. The status is: 1 is 1. Use, then enumerate the state, choose two different particles that are not used to try to collide

dp equations are dp [i | (1 << k)] = max (dp [i | (1 << k)], dp [i] + a [j] [k]), where I represents the state,, I represent status,,, I represent status,, I represent status,, I represent status,, I represent status,,, I represent status,, I represent status,, I represent status,, I represent status,,, I represent status,, I represent status. K indicates the latter particles of collision, J means the first particles of the collision

``````#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n, a;
int dp[1 << 11];

int main()
{
while(scanf("%d", &n) != EOF && n)
{
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
scanf("%d", &a[i][j]);
memset(dp, 0, sizeof(dp));
for(int i = 0; i < (1 << n); i++)
for(int j = 0; j < n; j++)
if(!(i & (1 << j)))
for(int k = 0; k < n; k++)
if(j != k && !(i & (1 << k)))
dp[i | (1 << k)] = max(dp[i | (1 << k)], dp[i] + a[j][k]);
int ans = 0;
for(int i = 0; i < (1 << n); i++)
ans = max(ans, dp[i]);
printf("%d\n", ans);
}
}``````

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