1105. Spiral Matrix (25)

时间限制
150 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 104. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:

12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 93
42 37 81
53 20 76
58 60 76

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注意控制好移动轨迹。
ni,nj是下次准备移动的目标点，若目标的越界或者已经赋值，则转向。

if(b[ni][nj]==-1&&ni<m&&nj<n&&ni>=0&&nj>=0){            i=ni,j=nj;            b[i][j]=a[cnt];            ni+=w[c%4][0],nj+=w[c%4][1];            cnt++;        }        else{            c++;            ni=i+w[c%4][0],nj=j+w[c%4][1];        }

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#include<iostream>#include<algorithm>#include<vector>using namespace std;int w[4][2]={{0,1},{1,0},{0,-1},{-1,0}};bool cmp(int a,int b){    return a>b;}int main(){    freopen("in.txt","r",stdin);    int N,x,b[110][110];    fill(b[0],b[0]+110*110,-1);    scanf("%d",&N);    vector<int> a(N);    for(int i=0;i<N;i++){        scanf("%d",&a[i]);    }    sort(a.begin(),a.end(),cmp);    int m=N,n=1;    for(int i=1;i<=N;i++){        if(N%i==0&&i>=(N/i)){            m=i,n=N/i;            break;        }    }    int i,j,cnt=0,c=0,ni=0,nj=0;    while(cnt<N){        if(b[ni][nj]==-1&&ni<m&&nj<n&&ni>=0&&nj>=0){            i=ni,j=nj;            b[i][j]=a[cnt];            ni+=w[c%4][0],nj+=w[c%4][1];            cnt++;        }        else{            c++;            ni=i+w[c%4][0],nj=j+w[c%4][1];        }    }    for(i=0;i<m;i++){        printf("%d",b[i][0]);        for(j=1;j<n;j++){            printf(" %d",b[i][j]);        }        printf("\n");    }    return 0;}