ZOJ1051 A New Growth Industry

這道題就是讀懂題目太費勁了，非要扯到什麼DNA上去，其實就是簡單的數組處理，和上下左右的加加，然後查表把所得值加到當前項上來，若越界則處理下。

#include <iostream>
using namespace std;

const int  MAXNUM = 20;//培養皿是*20的大小
char SignTable[]=".!X#";//符號表  
int dish[MAXNUM][MAXNUM],res[MAXNUM][MAXNUM];   
int day,d[16];   

int main()   
{    
    int cases;//測試範例數  
    int i,j,k;  
    while (cin>>cases)
    {
        while (cases--)
        {   
            cin>>day; //培養天數
            //輸入DNA序列資訊
            for (k=0; k<16; ++k)
                cin>>d[k];
            //輸入培養皿資料
            for (i=0; i<MAXNUM; ++i)   
                for (j=0; j<MAXNUM; ++j)   
                    cin>>dish[i][j];   
            while (day--)
            {   
                for (i=0; i<MAXNUM; ++i)   
                    for (j=0; j<MAXNUM; ++j)
                    {   
                        k = dish[i][j];  
                        //和上下左右的結合起來
                        if (i-1>=0) 
                            k += dish[i-1][j];   
                        if (i+1<MAXNUM) 
                            k += dish[i+1][j];   
                        if (j-1>=0)
                            k += dish[i][j-1];   
                        if (j+1<MAXNUM) 
                            k += dish[i][j+1];   
                        res[i][j] = dish[i][j]+d[k];   
                        //不能超過0~3的範圍
                        if (res[i][j]>3)
                            res[i][j] = 3;   
                        if (res[i][j]<0)
                            res[i][j] = 0;   
                    }   
                memcpy (dish,res,sizeof(dish));   
            }   
            for (i=0; i<MAXNUM; ++i)
            {   
                for (j=0; j<MAXNUM; ++j)   
                    cout<<SignTable[dish[i][j]];   
                cout<<endl;   
            }   
            //範例之間有一個空行
            if (cases!=0) 
                cout<<endl;   
        } 
    }
    return 0;   
}