noip模拟赛 蒜头君的坐骑

分析：标准的棋盘dp问题.

如果没有技能，那么就很好做了，相当于传纸条的做法.有了技能的限制，我们就要加上一维表示用了多少次技能，这个时候转移就要用到dfs了，而且不能用填表法，要用刷表法，从当前位置用技能的状态来更新到达的位置的状态,dp状态转移方程也要写成刷表法的形式.

如果很难从以前的状态推得现在的状态，那么就用现在的状态去更新以后的状态吧！

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
const int inf = 0x7ffffff;

int n, m, t, k, h, atk,f[][][],a[][],ans = inf;

int jisuan(int xue, int gongji)
{
    return(h - ) / gongji * xue;
}

void dfs(int x,int y,int cur,int xue,int gongji,int depth)
{
    if (depth > k)
    {
        f[x][y][cur + ] = min(f[x][y][cur + ], xue);
        return;
    }
    if (x +  <= n)
        dfs(x + , y, cur, xue + jisuan(a[x + ][y],gongji + a[x + ][y]), gongji + a[x + ][y], depth + );
    if (y +  <= m)
        dfs(x, y + , cur, xue + jisuan(a[x][y + ], gongji + a[x][y + ]), gongji + a[x][y + ], depth + );
}

int main()
{
    scanf("%d%d%d%d%d%d", &n, &m, &t, &k, &h, &atk);
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++)
            scanf("%d", &a[i][j]);
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++)
            for (int l = ; l <= t; l++)
                f[i][j][l] = inf;
    f[][][] = ;
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++)
            for (int l = ; l <= t; l++)
            {
                if (l != t)
                    dfs(i, j, l, f[i][j][l], atk,);
                if (i +  <= n)
                    f[i + ][j][l] = min(f[i + ][j][l], f[i][j][l] + jisuan(a[i + ][j], atk));
                if (j +  <= m)
                    f[i][j + ][l] = min(f[i][j + ][l], f[i][j][l] + jisuan(a[i][j + ], atk));
            }
    for (int i = ; i <= t; i++)
        ans = min(ans, f[n][m][i]);
    printf("%d\n", ans);

    return ;
}