[简单DP] COJ 1122 Game

这道题直接模拟即可，DP相当于一个优化，记录下从第i行第j列出发后到达的最终位置，下次需要时直接取，优化前后最坏情况下复杂度都是O(10^6)。

# include <cstdio>

# include <cstring>

 

# define N 1000 + 5

 

int n, m;

int cnt[N], ff[N], f[N][N];

char g[N][N];

 

int down(int row, int col)

{

    int &ans = f[row][col];

    if (ans != -1) return ans;

    if (row == n) return ans = col;

    if (g[row][col] == '\\')

    {

        if (col+1>m || g[row][col+1]=='/') return ans = 0;

        return ans = down(row+1, col+1);

    }

    if (g[row][col] == '/')

    {

        if (col-1<1 || g[row][col-1]=='\\') return ans = 0;

        return ans = down(row+1, col-1);

    }

    return ans = down(row+1, col);

}

 

void solve(void)

{

    for (int i = 1; i <= n; ++i)

        memset(f[i]+1, -1, sizeof(int)*m);   

    for (int i = 1; i <= m; ++i)

    {

        ff[i] = down(1, i);

        cnt[i] = 0;

    }

    int k, x;

    scanf("%d", &k);

    while (k--)

    {

        scanf("%d", &x);

        ++cnt[ff[x]];

    }

    printf("%d", cnt[1]);

    for (int i = 2; i <= m; ++i)

        printf(" %d", cnt[i]);

    putchar('\n');

}

 

void read_graph(void)

{

    for (int i = 1; i <= n; ++i)

        scanf("%s", g[i]+1);

}

 

int main()

{

    while (~scanf("%d%d", &n, &m))

    {

        read_graph();

        solve();

    }   

     

    return 0;

}

2WA，输出时把 n 当作 m 了，还以为是题目的下落方式没理解对。

一组数据：

------------------------input--------------------------------
5 5
_____
_\\__
__/__
_\___
_____
5 1 2 3 4 5

------------------------output------------------------------
1 0 1 2 1