HDU 2258 Continuous Same Game

HDU_2258

    本来是Step里面遇到了HDU_2259的，结果发现要先做这个题……模拟题目中所说的规则就可以了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 25
#define MAXD 410
int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1}, L;
int N, M, a[MAXN][MAXN];
struct List
{
    int x, y;
}list[MAXD];
void init()
{
    int i, j;
    char b[MAXN];
    for(i = N - 1; i >= 0; i --)
    {
        scanf("%s", b);
        for(j = 0; j < M; j ++) a[i][j] = b[j] - '0';
    }
}
struct St
{
    int N, M, g[MAXN][MAXN], cnt[MAXD], col[MAXN][MAXN];
    void init(int N, int M, int g[][MAXN])
    {
        this->N = N, this->M = M;
        memcpy(this->g, g, sizeof(this->g));
    }
    inline int inside(int x, int y)
    {
        return x >= 0 && x < N && y >= 0 && y < M;
    }
    void color(int x, int y, int &cnt, int c)
    {
        int i, nx, ny;
        col[x][y] = c, ++ cnt;
        for(i = 0; i < 4; i ++)
        {
            nx = x + dx[i], ny = y + dy[i];
            if(inside(nx, ny) && g[nx][ny] == g[x][y] && !col[nx][ny])
                color(nx, ny, cnt, c);
        }
    }
    void erase(int x, int y)
    {
        int i, nx, ny;
        g[x][y] = 0;
        for(i = 0; i < 4; i ++)
        {
            nx = x + dx[i], ny = y + dy[i];
            if(inside(nx, ny) && col[x][y] == col[nx][ny] && g[nx][ny])
                erase(nx, ny);
        }
    }
    inline int find(int j)
    {
        for(int i = 0; i < N; i ++) if(g[i][j]) return 0;
        return 1;
    }
    void arrange()
    {
        int i, j, k, kmax;
        kmax = 0;
        for(j = 0; j < M; j ++)
        {
            k = 0;
            for(i = 0; i < N; i ++)
                if(g[i][j])
                {
                    if(i != k)
                        g[k][j] = g[i][j], g[i][j] = 0;
                    ++ k;
                }
            kmax = std::max(kmax, k);
        }
        N = kmax;
        k = 0;
        for(j = 0; j < M; j ++)
            if(!find(j))
            {
                if(j != k) for(i = 0; i < N; i ++) g[i][k] = g[i][j];
                ++ k;
            }
        M = k;
    }
    int gogo()
    {
        int i, j, ans = 0, max, x, y, c;
        for(;;)
        {
            max = c = 0;
            for(i = 0; i < N; i ++)
                for(j = 0; j < M; j ++) col[i][j] = 0;
            for(i = N - 1; i >= 0; i --)
                for(j = 0; j < M; j ++)
                    if(g[i][j] && !col[i][j])
                    {
                        ++ c, cnt[c] = 0;
                        color(i, j, cnt[c], c);
                        if(cnt[c] > max) max = cnt[c], x = i, y = j;
                    }
            if(max <= 1) break;
            ans += (max - 1) * max;
            erase(x, y), arrange();
        }
        return ans;
    }
};
void solve()
{
    St st;
    st.init(N, M, a);
    printf("%d\n", st.gogo());
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();
    }
    return 0;
}
/*
Sample Input:
    
5 5
32345
32345
32345
32345
12111

5 5
32223
32213
32223
32221
12111

Sample Output:

112
204

*/