ZOJ1051 A New Growth Industry
2008-10-23 19:45
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这道题就是读懂题目太费劲了,非要扯到什么DNA上去,其实就是简单的数组处理,和上下左右的加加,然后查表把所得值加到当前项上来,若越界则处理下。
#include <iostream>
using namespace std;
const int MAXNUM = 20;//培养皿是*20的大小
char SignTable[]=".!X#";//符号表
int dish[MAXNUM][MAXNUM],res[MAXNUM][MAXNUM];
int day,d[16];
int main()
{
int cases;//测试样例数
int i,j,k;
while (cin>>cases)
{
while (cases--)
{
cin>>day; //培养天数
//输入DNA序列信息
for (k=0; k<16; ++k)
cin>>d[k];
//输入培养皿数据
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];
//和上下左右的结合起来
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];
//不能超过0~3的范围
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;
}
//样例之间有一个空行
if (cases!=0)
cout<<endl;
}
}
return 0;
}
#include <iostream>
using namespace std;
const int MAXNUM = 20;//培养皿是*20的大小
char SignTable[]=".!X#";//符号表
int dish[MAXNUM][MAXNUM],res[MAXNUM][MAXNUM];
int day,d[16];
int main()
{
int cases;//测试样例数
int i,j,k;
while (cin>>cases)
{
while (cases--)
{
cin>>day; //培养天数
//输入DNA序列信息
for (k=0; k<16; ++k)
cin>>d[k];
//输入培养皿数据
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];
//和上下左右的结合起来
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];
//不能超过0~3的范围
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;
}
//样例之间有一个空行
if (cases!=0)
cout<<endl;
}
}
return 0;
}
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