NOI 2005 瑰丽华尔兹（三维DP + 单调队列优化）

思路：

1. dp[k][x][y] 表示处理到第 k 个时间区间时，最终点落在 x, y 坐标上，移动的最大距离。

3. dp[k][x][y] = max(dp[k-1][x1][y1] + delta); 由于 delta 是相对偏移量，所以对于 x/y 所在的维度可以用单调队列优化。

4. 由于第 k - 1 次之后，无法确定第 k 步的起始位置，所以要采取枚举的办法，所以最终的时间复杂度为 O(N*M*K)

5. 初始状态为 dp[0][x][y] = 0, 其他赋值为 -INFS，这样才能保证枚举的过程中，最终结果是在以 (x, y) 为起始点出发的。

#include <iostream>

#include <cstdio>

#include <algorithm>

using namespace std;

 

const int MAXN = 210;

const int INFS = 0x3fffffff;

 

int dx[5] = {-1, 1, 0, 0};

int dy[5] = {0, 0, -1, 1};

 

char grid[MAXN][MAXN];

int t1, t2, dp[2][MAXN][MAXN], deq[MAXN], pos[MAXN];

 

void solvedp(int x, int y, int width, int size, int d)

{

    int s = 0, e = -1;

    for (int i = 1; i <= width; ++i)

    {

        if (grid[x][y] == '.')

        {

            int val = dp[t1][x][y] - i;

            while (s <= e && deq[e] < val)

                --e;

 

            deq[++e] = val, pos[e] = i;

            while (i - pos[s] > size)

                ++s;

 

            dp[t2][x][y] = deq[s] + i;

        }

        else

        {

            s = 0, e = -1;

            dp[t2][x][y] = -INFS;

        }

        x += dx[d], y += dy[d];

    }

}

 

int main()

{

    int N, M, x, y, K;

    while (scanf("%d %d %d %d %d", &N, &M, &x, &y, &K) != EOF)

    {

        for (int i = 1; i <= N; ++i)

            scanf("%s", &grid[i][1]);

 

        for (int i = 1; i <= N; ++i)

            for (int j = 1; j <= M; ++j)

                dp[0][i][j] = -INFS;

        dp[0][x][y] = 0;

 

        t1 = 1, t2 = 0;

        for (int i = 0; i < K; ++i)

        {

            int si, ti, di;

            scanf("%d %d %d", &si, &ti, &di);

 

            t1 ^= 1, t2 ^= 1;

             

            if (di == 1)

            {

                for (int j = 1; j <= M; ++j)

                    solvedp(N, j, N, ti - si + 1, di - 1);

            }

            else if (di == 2)

            {

                for (int j = 1; j <= M; ++j)

                    solvedp(1, j, N, ti - si + 1, di - 1);

            }

            else if (di == 3)

            {

                for (int j = 1; j <= N; ++j)

                    solvedp(j, M, M, ti - si + 1, di - 1);

            }

            else if (di == 4)

            {

                for (int j = 1; j <= N; ++j)

                    solvedp(j, 1, M, ti - si + 1, di - 1);

            }

        }

 

        int ans = 0;

        for (int i = 1; i <= N; ++i)

            for (int j = 1; j <= M; ++j)

                ans = max(ans, dp[t2][i][j]);

 

        printf("%d\n", ans);

    }

    return 0;

}