小p的2048

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题目大意:2048游戏规则如下:balbalbala...但是小p为了降低难度,把规则改了一下,具体呢就是给定你操作的方向和新生成的2或者4的位置,问你最后结束的时候有效操作了几次和最高得分(每次合并的分数加起来)

我是2048的专业户...

所以做这个题只要稍微联系一个游戏就好了,然后就没了.(我代码玄学90pts,如果后续有学弟愿意深究这个题的话,不介意你们帮我D一下)

代码如下: 

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#include <bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;

inline ll read()
{
    ll s(0), w(1);
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-') w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return s * w;
}

inline void put(ll x)
{
    if (! x) putchar('0');
    if (x < 0) putchar('-'), x = -x;
    int num(0); char c[66];
    while (x) c[++ num] = x % 10 + 48, x /= 10;
    while (num) putchar(c[num --]);
    return (void)(putchar('\n'));
}

const int N = 66;

int n, m, sx, sy, sv, tx, ty, tv;
int D, K, V, opt, res, SB1, SB2;
int a[N][N], b[N][N];
bool vis[N][N];

inline void yhm_Born()
{
    int i, j, cnt(0), r(0);
    for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j) if (! a[i][j]) ++ r;
    int who = (1 + K % r);
    for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j) if (! a[i][j])
    {
        ++ cnt;
        if (cnt == who)
        {
            a[i][j] = V;
            SB1 = i, SB2 = j;
            break;
        }
    }
    return;
}

inline void yhm_up()
{
    int i, j; memset(vis, 0, sizeof vis);
    for (j = 1; j <= n; ++ j)
    {
        for (i = 2; i <= n; ++ i)
        {
            int cnt = i;
            if (! a[i][j]) continue;
            while (cnt > 1 && a[cnt - 1][j] == 0) -- cnt;//跳到空格
            if (cnt == i);
            else a[cnt][j] = a[i][j], a[i][j] = 0;
            if (a[cnt - 1][j] == a[cnt][j] && ! vis[cnt - 1][j])
                a[cnt - 1][j] *= 2, res += a[cnt - 1][j], a[cnt][j] = 0, vis[cnt - 1][j] = 1;
        }
    }
    return;
}

inline void yhm_dn()
{
    int i, j; memset(vis, 0, sizeof vis);
    for (j = 1; j <= n; ++ j)
    {
        for (i = n - 1; i >= 1; -- i)
        {
            int cnt = i;
            if (! a[i][j]) continue;
            while (cnt < n && a[cnt + 1][j] == 0) ++ cnt;//跳到空格
            if (cnt == i);
            else a[cnt][j] = a[i][j], a[i][j] = 0;
            if (a[cnt + 1][j] == a[cnt][j] && ! vis[cnt + 1][j])
                a[cnt + 1][j] *= 2, res += a[cnt + 1][j], a[cnt][j] = 0, vis[cnt + 1][j] = 1;
        }
    }
    return;
}

inline void yhm_lt()
{
    int i, j; memset(vis, 0, sizeof vis);
    for (i = 1; i <= n; ++ i)
    {
        for (j = 2; j <= n; ++ j)
        {
            int cnt = j;
            if (! a[i][j]) continue;
            while (cnt > 1 && a[i][cnt - 1] == 0) -- cnt;
            if (cnt == j);
            else a[i][cnt] = a[i][j], a[i][j] = 0;
            if (a[i][cnt - 1] == a[i][cnt] && ! vis[i][cnt - 1])
                a[i][cnt - 1] *= 2, res += a[i][cnt - 1], a[i][cnt] = 0, vis[i][cnt - 1] = 1;
        }
    }
    return;
}

inline void yhm_rt()
{
    int i, j; memset(vis, 0, sizeof vis);
    for (i = 1; i <= n; ++ i)
    {
        for (j = n - 1; j >= 1; -- j)
        {
            int cnt = j;
            if (! a[i][j]) continue;
            while (cnt < n && a[i][cnt + 1] == 0) ++ cnt;
            if (cnt == j);
            else a[i][cnt] = a[i][j], a[i][j] = 0;
            if (a[i][cnt + 1] == a[i][cnt] && ! vis[i][cnt + 1])
                a[i][cnt + 1] *= 2, res += a[i][cnt + 1], a[i][cnt] = 0, vis[i][cnt + 1] = 1;
        }
    }
    return;
}

inline void yhm_Func()
{
    if (D == 0) return (void)(yhm_up());
    if (D == 1) return (void)(yhm_dn());
    if (D == 2) return (void)(yhm_lt());
    if (D == 3) return (void)(yhm_rt());
    return;
}

inline bool yhm_Check()
{
    int i, j, pd(0);
    for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j)
    {
        if (i == SB1 && j == SB2) continue;
        if (a[i][j] == b[i][j]) ++ pd;
    }
    if (pd == n * n - 1) return true;
    return false;
}

inline void yhm_chenge()
{
    int i, j;
    for (i = 1; i <= n; ++ i) for (j = 1; j <= n; ++ j) b[i][j] = a[i][j];
    return;
}

signed main()
{
    n = read(), m = read();
    sx = read(), sy = read(), sv = read(), tx = read(), ty = read(), tv = read();
    a[sx][sy] = sv, a[tx][ty] = tv;

    while (m --)
    {
        D = read(), K = read(), V = read();
        yhm_Func();
        if (yhm_Check())
        {
            put(opt), put(res);
            return 0;
        }
        yhm_Born();
        ++ opt;
        yhm_chenge();

    }
    
    put(opt), put(res);
    return 0;
}

后记:

这是继红心大战之后又一道模拟题

2048的原型就很好,自己的码力还是可以的,要兢兢业业的保持到noip!