poj 2078 Matrix

最新推荐文章于 2020-07-02 17:44:41 发布

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本文介绍了一个关于矩阵操作的问题：通过任意次数地将矩阵的某一行元素向右移动一位，并让最右端的元素移至最左端的方式，来最小化各列元素之和的最大值。文章提供了一种搜索算法实现思路及完整代码。

Description
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
max0<=j< n{Cj|Cj=Σ0<=i< nAi,j}

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 |Ai,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

11
15
好久没写搜索了额，总算是写出来了吧！

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
int a[10][10];
int n,mini,maxi;
void shift(int k){
    int kk=a[k][n-1];
    for(int i=n-1;i>=1;i--)
        a[k][i]=a[k][i-1];
        a[k][0]=kk;
}
void dfs(int co){             对行搜索
    if(co==n)
        return ;
        maxi=-9999999;
        for(int ii=0;ii<n;ii++){
                int sum=0;
            for(int jj=0;jj<n;jj++)
                 sum+=a[jj][ii];
                maxi=max(sum,maxi);
        }
        mini=min(maxi,mini);
        for(int ii=0;ii<n;ii++){ 对每行shift n-1次
            shift(co+1);
            dfs(co+1);
        }
}
int main(){
    while(scanf("%d",&n)!=EOF&&n!=-1){
        mini=99999999,maxi=-99999999;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&a[i][j]);
         dfs(0);
         printf("%d\n",mini);
    }
    return 0;
}