在生命矩阵游戏中搜索人口最多的区域

Question

我用C语言编写了一个生活游戏，它运行得很好，唯一的问题是游戏的棋盘比显示的要大得多。我知道显示器可以显示的行数和列数，我认为跟踪操作(搜索人口最多的区域)是一个很好的想法。因此，我写了一个小函数来接收我应该显示的部分字段的角的4个索引，但不知何故代码似乎给出了随机部分，这有时显然是假的，活细胞(max)的计算值完全是假的(有时甚至比整个矩阵本身还要大)。代码如下：

int* most_populated_area(int m, int n, bool a[m][n], int r, int c){
    int * rr=malloc(sizeof(int)*4); //return value
    rr[0]=0;
    rr[1]=r;
    rr[2]=0;
    rr[3]=c; //we first assume that the most populated arrea is upper left corner
    int summe=0;
    int max=0; //to controll wheater or not we found a more populated area we need a summe and the max we founded until now
    for(int isearch=0; isearch<m; isearch++){ //we search the matrix row for row
        for(int jsearch=0; jsearch<n; jsearch++){ // and column for column
            for(int ik=0; ik<=r; ik++){ // now from the current flied we go as many rows we are allowed
                for(int jl=0; jl<=c; jl++){ //and also as many columns we are allowed to go
                    int iks=isearch+ik; //we looking on the field is 0 to r
                    int jls=jsearch+jl; //and 0 to c flieds away from our current field
                    if(iks>=m){ //if the field is outside the matrix we have to go back  
                        iks%=r; // iks is clearly bigger than r (because m bigger than r) but iks%r can be r-1 at most,
                        //so we have the rest from iks through r over the edge 
                    }
                    if(jls>=n){ //analog to iks
                        jls%=c;
                    }
                    summe+=a[iks][jls];
                }
            }
            if(summe>max){// the summe is greater than the max we found a more populated area and we save the the results
                max=summe;
                rr[0]=isearch;
                rr[1]=(isearch+r>=m)?(isearch+r)%r:isearch+r;
                rr[2]=jsearch;
                rr[3]=(jsearch+c>=n)?(jsearch+c)%c:jsearch+c;
            }
            summe=0; //anyway we reset the summe to search accurate
        }
    }
    printf("Lifing in this area %d \n", max);
    return rr;
}

Answer 1

有一个近似解是O(n)。

将电路板拆分为x和y分量，并搜索最多的X和Y值。然后将它们设置为要显示的中点(针对边进行了调整)。