uva-387-暴力枚举

题意:

给你一些小方块,问是不是能组成一个4X4的大方块,所有方块全部要使用,裸枚举

#include <iostream>
#include <stdio.h>
#include <memory.h>

using namespace std;
const int NN = 20;
class Piece
{
    private:
        int r;
        int c;
        int** p;
    public:
        Piece()
                : r(0), c(0)
        {
            p = NULL;
        }
        Piece(int r, int c, int (*pp)[NN])
        {
            this->r = r;
            this->c = c;
            this->p = new int*[r];

            for(int i = 0; i < r; i++)
            {
                this->p[i] = new int[c];
                memcpy(this->p[i], pp[i], c * sizeof(int));
            }
        }
        ~Piece()
        {
            for(int i = 0; i < r; i++)
                delete[] p[i];
            delete this->p;
        }
        int getR()
        {
            return this->r;
        }
        int getC()
        {
            return this->c;
        }
        int ** getP()
        {
            return this->p;
        }

};

int N;
int r, c;
int m[6][6];
int vis[NN];
Piece* piece[NN];
int pi = 0;
void dump()
{
    for(int i = 0; i < pi; i++)
        for(int j = 0; j < piece[i]->getR(); j++)
        {
            for(int k = 0; k < piece[i]->getC(); k++)
                cout << piece[i]->getP()[j][k];
            cout << endl;
        }
}
bool judge(int y, int x, Piece* p)
{
    for(int i = 0; i < p->getR(); i++)
        for(int j = 0; j < p->getC(); j++)
            if(p->getP()[i][j] != 0 && m[i + y][x + j] != 0)
            {
                return false;
            }
    return true;
}
void reset(int x, int y, int r, int c, int cur)
{
    for(int i = 0; i < r; i++)
        for(int j = 0; j < c; j++)
        {
            if(m[y + i][x + j] != cur)
                continue;
            m[y + i][x + j] = 0;
        }
}

void copy(int x, int y, int r, int c, int** src)
{
    for(int i = 0; i < r; i++)
        for(int j = 0; j < c; j++)
        {
            if(m[y + i][x + j])
                continue;
            m[y + i][x + j] = src[i][j];
        }
}
bool dfs(int cur)
{
    if(cur == pi)
    {
        for(int i = 0; i < 4; i++)
        {
            for(int j = 0; j < 4; j++)
                if(m[i][j] == 0)
                    return false;
        }
        return true;
    }
    //对当前的cur,枚举每一个坐标X,Y
    for(int i = 0; i < 4; i++)
    {
        for(int j = 0; j < 4; j++)
        {
            if(m[i][j] && (piece[cur]->getP()[0][0]))
                continue;
            if(!judge(i, j, piece[cur]))
                continue;
            int r = piece[cur]->getR();
            int c = piece[cur]->getC();
            copy(j, i, r, c, piece[cur]->getP());
            int ok = dfs(cur + 1);
            if(ok)
                return ok;
            reset(j, i, r, c, cur + 1);
        }
    }
    return false;
}
int main()
{
    //freopen("d://1.text", "r", stdin);
    int t = 0;
    while (cin >> N && N)
    {
        if(t != 0)
            cout << endl;
        ++t;
        memset(m, 0, sizeof(m));
        memset(piece, 0, sizeof(piece));
        memset(vis, 0, sizeof(vis));
        int p[NN][NN];
        pi = 0;
        for(int i = 1; i <= N; i++)
        {
            scanf("%d %d", &r, &c);
            for(int j = 0; j < r; j++)
                for(int k = 0; k < c; k++)
                {
                    char t;
                    cin >> t;
                    if(t == '0')
                        p[j][k] = 0;
                    else
                        p[j][k] = i;
                }
            Piece* pp = new Piece(r, c, p);
            piece[pi++] = pp;
        }
        bool ok = dfs(0);
        if(!ok)
        {
            cout << "No solution possible" << endl;
        }
        else
        {
            for(int i = 0; i < 4; i++)
            {
                for(int j = 0; j < 4; j++)
                    cout << m[i][j];
                cout << endl;
            }
        }
    }
//    dump();

    return 0;
}
原文地址:https://www.cnblogs.com/shuiyonglewodezzzzz/p/9221362.html