Codeforces Round #487 (Div. 2) C. A Mist of Florescence 构造

题意：

让你构造一个 n∗mn*mn∗m 矩阵，这个矩阵由 444 种字符填充构成，给定 444 个整数，即矩阵中每种字符构成的联通块个数，n,mn,mn,m 需要你自己定，但是不能超过505050.

数据范围：

每个字符组成的联通块个数不超过100.

题解：

正着生成联通块似乎有些费劲，不如我们逆着思考问题将400个联通块生成，并逐渐减少联通块的个数。

A picture is worth a thousand words

任意发挥想象吧！

Code:

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 100;
char str[maxn][maxn];
inline void init()
{
    for(int i = 1;i <= 25; ++i)
        for(int j = 1;j <= 25; ++j) str[i][j] = 'B';
    for(int i = 1;i <= 25; ++i)
        for(int j = 26;j <= 50;++j) str[i][j] = 'A';

    for(int i = 26;i <= 50; ++i)
        for(int j = 1;j <= 25; ++j) str[i][j] = 'C';
    for(int i = 26;i <= 50; ++i)
        for(int j = 26;j <= 50; ++j) str[i][j] = 'D';
    int cnt = 0;
    for(int i = 2;i < 25; i += 2)
    {
        for(int j = 2;j < 25; j += 2)
        {
            if(cnt >= 99) break;
            ++cnt;
            str[i][j] = 'A';
        }
    }
    cnt = 0;
    for(int i = 2;i < 25; i += 2)
    {
        for(int j = 27;j < 50;j += 2)
        {
            if(cnt >= 99) break;
            ++cnt;
            str[i][j] = 'B';
        }
    }
    cnt = 0;
    for(int i = 27;i < 50;i += 2)
        for(int j = 2;j < 25; j += 2)
        {
            if(cnt >= 99) break;
            ++cnt;
            str[i][j] = 'D';
        }
    cnt = 0;
    for(int i = 27;i < 50;i += 2)
        for(int j = 27;j < 50;j += 2)
        {
            if(cnt >= 99) break;
            ++cnt;
            str[i][j] = 'C';
        }

}
int main()
{
    //freopen("input.in","r",stdin);
    int a,b,c,d;
    scanf("%d%d%d%d",&a,&b,&c,&d);
    printf("50 50\n");
    init();
    a = 100 - a, b = 100 - b, c = 100 - c, d = 100 - d;
    int cnt_a = 0, cnt_b = 0, cnt_c = 0, cnt_d = 0;
    for(int i = 2;i < 25; i += 2)
        for(int j = 2;j < 25; j += 2)
        {
            if(cnt_a >= a) break;
            ++cnt_a;
            str[i][j] = 'B';
        }
    for(int i = 2;i < 25; i += 2)
        for(int j = 27;j < 50;j += 2)
        {
            if(cnt_b >= b) break;
            ++cnt_b;
            str[i][j] = 'A';
        }
    for(int i = 27;i < 50;i += 2)
        for(int j = 2;j < 25; j += 2)
        {
            if(cnt_d >= d) break;
            ++cnt_d;
            str[i][j] = 'C';
        }
    for(int i = 27;i < 50;i += 2)
        for(int j = 27;j < 50;j += 2)
        {
            if(cnt_c >= c) break;
            ++cnt_c;
            str[i][j] = 'D';
        }
    for(int i = 1;i <= 50; ++i)
    {
        for(int j = 1;j <= 50; ++j) printf("%c",str[i][j]);
        printf("\n");
    }
    return 0;

}