ZOJ 2418 Matrix

最新推荐文章于 2019-09-19 15:49:50 发布

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最新推荐文章于 2019-09-19 15:49:50 发布

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分类专栏： ZOJ ----dfs 文章标签： zoj

本文链接：https://blog.youkuaiyun.com/jtjy568805874/article/details/70239309

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本文介绍了一个特定问题：通过任意次数地将矩阵的某一行元素右移一位来最小化矩阵列和的最大值。文章提供了完整的C++代码实现，并解释了如何通过遍历所有可能的操作组合来寻找最优解。

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Given an n*n matrix A, whose entries Ai,j are integer numbers ( 0 <= i < n, 0 <= j < n ). An operation SHIFT at row i ( 0 <= i < n ) will move the integers in the row one position right, and the rightmost integer will wrap around to the leftmost column.

You can do the SHIFT operation at arbitrary row, and as many times as you like. Your task is to minimize

Input

The input consists of several test cases. The first line of each test case contains an integer n. Each of the following n lines contains n integers, indicating the matrix A. The input is terminated by a single line with an integer -1. You may assume that 1 <= n <= 7 and |A_i,j| < 10⁴.

Output

For each test case, print a line containing the minimum value of the maximum of column sums.

Sample Input

2
4 6
3 7
3
1 2 3
4 5 6
7 8 9
-1

Sample Output

直接爆搜就可以过。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
#define rep(i,j,k) for (int i=j;i<=k;i++)
#define per(i,j,k) for (int i=j;i>=k;i--)
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
typedef pair<int,int> pii;
const int N = 15;
const int INF=0x7FFFFFFF;
int n,a[N][N],b[N],ans;

int calc()
{
    int res=-INF;
    rep(i,0,n-1)
    {
        int cnt=0;
        rep(j,0,n-1)
        {
            cnt+=a[j][(i+b[j])%n];
        }
        res=max(res,cnt);
    }
    return res;
}

void dfs(int x)
{
    if (x==n)
    {
        ans=min(ans,calc());
        return;
    }
    rep(i,0,n-1)
    {
        b[x]=i;
        dfs(x+1);
    }
}

int main()
{
    while (~inone(n)&&n!=-1)
    {
        rep(i,0,n-1) rep(j,0,n-1) inone(a[i][j]);
        ans=INF; b[0]=0; dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}