HDU 1057 模拟细菌的变化

最新推荐文章于 2017-09-09 13:37:03 发布

置顶 sky_miange

最新推荐文章于 2017-09-09 13:37:03 发布

阅读量941

点赞数

CC 4.0 BY-SA版权

分类专栏： HDU 水题文章标签： ACM算法 amp math.h printf c

本文链接：https://blog.youkuaiyun.com/sky_miange/article/details/44022401

HDU 水题专栏收录该内容

112 篇文章

订阅专栏

本文介绍了一个简单的细菌生长模拟程序，该程序根据预设的DNA规则和初始密度，在20x20的培养皿中模拟细菌群体随时间的变化情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A New Growth Industry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1372 Accepted Submission(s): 541

Problem Description

A biologist experimenting with DNA modification of bacteria has found a way to make bacterial colonies sensitive to the
surrounding population density. By changing the DNA, he is able to “program” the bacteria to respond to the varying densities in their immediate neighborhood.

The culture dish is a square, divided into 400 smaller squares (20x20). Population in each small square is measured on a four point scale (from 0 to 3). The DNA information is represented as an array D, indexed from 0 to 15, of integer values and is interpreted as follows:

In any given culture dish square, let K be the sum of that square's density and the densities of the four squares immediately to the left, right, above and below that square (squares outside the dish are considered to have density 0). Then, by the next day, that dish square's density will change by D[K] (which may be a positive, negative, or zero value). The total density cannot, however, exceed 3 nor drop below 0.

Now, clearly, some DNA programs cause all the bacteria to die off (e.g., [-3, -3, …, -3]). Others result in immediate population explosions (e.g., [3,3,3, …, 3]), and others are just plain boring (e.g., [0, 0, … 0]). The biologist is interested in how some of the less obvious DNA programs might behave.

Write a program to simulate the culture growth, reading in the number of days to be simulated, the DNA rules, and the initial population densities of the dish.

Input

Input to this program consists of three parts:

1. The first line will contain a single integer denoting the number of days to be simulated.

2. The second line will contain the DNA rule D as 16 integer values, ordered from D[0] to D[15], separated from one another by one or more blanks. Each integer will be in the range -3…3, inclusive.

3. The remaining twenty lines of input will describe the initial population density in the culture dish. Each line describes one row of squares in the culture dish, and will contain 20 integers in the range 0…3, separated from one another by 1 or more blanks.

Output

The program will produce exactly 20 lines of output, describing the population densities in the culture dish at the end of the simulation. Each line represents a row of squares in the culture dish, and will consist of 20 characters, plus the usual end-of-line terminator.

Each character will represent the population density at a single dish square, as follows:

No other characters may appear in the output.

Sample Input

  
   1

2 
0 1 1 1 2 1 0 -1 -1 -1 -2 -2 -3 -3 -3 -3 
3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sample Output

  
   ##!................. 
#!.................. 
!................... 
.................... 
.................... 
.................... 
.................... 
.........!.......... 
........!#!......... 
.......!#X#!........ 
........!#!......... 
.........!.......... 
.................... 
.................... 
.................... 
.................... 
.................... 
.................... 
.................... 
....................

Source

Mid-Atlantic USA 2001

题意：给定一个天数N，求20*20方阵内细菌的变化情况。每次变化都是加上一个d[k]，d数组有给定的16个数。k是值是它本身加上它上下左右的四个数。

简单模拟题。

#include <stdio.h>
#include <string.h>
int main()
{
    int d[16];
    int a[22][22];
    int b[22][22];
    int n,i,j,day,k;
    scanf("%d",&n);
    while(n--)
    {
        //if(n)
        //    printf("\n");
        memset(a,0,sizeof(a));
        scanf("%d",&day);
        memset(b,0,sizeof(b));
        for(i=0;i<16;i++)
            scanf("%d",&d[i]);
        for(i=1;i<=20;i++)
            for(j=1;j<=20;j++)
                scanf("%d",&a[i][j]);
            while(day--)
            {
                for(i=1;i<=20;i++)
                    for(j=1;j<=20;j++)
                    {
                        b[i][j]=a[i][j]+d[a[i-1][j]+a[i+1][j]+a[i][j-1]+a[i][j+1]+a[i][j]];//每次变化的值
                        if(b[i][j]>3)
                            b[i][j]=3;  //最大为3,
                        else if(b[i][j]<0)
                            b[i][j]=0;  //最小为0
                    }
                    for(i=1;i<=20;i++)
                        for(j=1;j<=20;j++)
                            a[i][j]=b[i][j]; //变化一次后重新赋值
            }
            for(i=1;i<=20;i++)
            {
                for(j=1;j<=20;j++)
                {
                    if(b[i][j]==0)
                        printf(".");
                    else if(b[i][j]==1)
                        printf("!");
                    else if(b[i][j]==2)
                        printf("X");
                    else
                        printf("#"); //输出。
                }
                printf("\n");
            }
            if(n)
            printf("\n");
    }
    return 0;
}