+Bessie+the+cow+and+her+friend+Elsie+like+to+play+a+version+of+a+game+t

Bessie the cow and her friend Elsie like to play a version of a game called "Tag". In their version, Bessie starts as the "tagger" and tries to catch Elsie. Once Bessie catches Elsie, they switch roles and Elsie becomes the new tagger. They enjoy playing this game to have fun and stay active.

相关问题

To+pass+the+time,+Bessie+the+cow+and+her+friend+Elsie+like+to+play+a+version+of+a+game+they+saw+at+t

To pass the time, Bessie the cow and her friend Elsie like to play a version of a game they saw at the county fair. The game involves throwing rings onto bottles to win prizes. Bessie and Elsie have set up their own version of the game in the barnyard using hula hoops and milk bottles. They take turns tossing the hula hoops and trying to land them on top of the milk bottles. It's a fun and entertaining way for them to spend their time together.

c++Bessie and friends have been captured and are trapped in a secret compound in a location far from their farm, and it is up to Bessie to plan their escape! The compound consists of � � NK holding cells arranged in an � × � N×K rectangular grid, where there are gates between horizontally and vertically adjacent cells. Each cell houses exactly one cow. Bessie has hacked into the system, and is able to unlock any subset of the gates, but each gate has a cost. For the cows to escape, Bessie must open enough gates that all the cows can gather in a single cell (so that they have enough cow-power to tunnel to the surface!). Bessie wants to minimize the total unlocking cost. But the stakes are higher than ever, and Bessie cannot be content with just one escape plan: she needs back-ups. Help her count the number of minimum-cost escape plans; two plans are considered different if some gate needs to be unlocked in one of the plans but not the other. Since this number may be very large, only output its remainder modulo 1 0 9 + 7 10 9 +7.

你好！根据题目描述，我们需要帮助 Bessie 计算最小成本逃生计划的数量，并且要对结果进行取模操作。为了解决这个问题，我们可以使用动态规划的方法来求解。

首先，我们定义一个二维数组 dp，其中 dp[i][j] 表示在位置 (i, j) 时的最小成本逃生计划数量。我们可以使用动态规划的思想进行状态转移。

根据题目要求，所有的牛必须聚集在同一个单元格中才能逃生，因此我们可以将问题分解为两个子问题：
1. 选择一个单元格 (i, j)，使得牛能够聚集在该单元格中。
2. 计算到达单元格 (i, j) 的最小成本逃生计划数量。

对于第一个子问题，假设我们选择了单元格 (i, j) 作为聚集点，那么该单元格一定有一个入口单元格 (x, y)，其中 (x, y) 是 (i, j) 的上方或左方单元格。因此，我们可以通过遍历所有可能的入口单元格来计算第二个子问题的结果。

对于第二个子问题，我们可以使用动态规划的方法进行求解。假设我们已经计算了 dp[x][y] 的结果，那么到达单元格 (i, j) 的最小成本逃生计划数量可以通过以下方式计算：
1. 如果 (i, j) 的上方单元格 (x, y) 存在，那么 dp[i][j] += dp[x][y]。
2. 如果 (i, j) 的左方单元格 (x, y) 存在，那么 dp[i][j] += dp[x][y]。

最后，我们需要遍历所有的单元格，找到其中最小成本逃生计划数量的最小值，并将结果对 10^9+7 取模。

下面是对应的 C++ 代码实现：

#include <iostream>
#include <vector>

using namespace std;

const int MOD = 1e9 + 7;

int countEscapePlans(vector<vector<int>>& gates) {
    int N = gates.size();
    int K = gates[0].size();

    vector<vector<int>> dp(N, vector<int>(K, 0));

    // 初始化边界条件
    dp[0][0] = 1;

    // 动态规划求解
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < K; j++) {
            if (i > 0) {
                dp[i][j] += dp[i - 1][j];
                dp[i][j] %= MOD;
            }
            if (j > 0) {
                dp[i][j] += dp[i][j - 1];
                dp[i][j] %= MOD;
            }
        }
    }

    return dp[N - 1][K - 1];
}

int main() {
    int N, K;
    cin >> N >> K;

    vector<vector<int>> gates(N, vector<int>(K, 0));

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < K; j++) {
            cin >> gates[i][j];
        }
    }

    int result = countEscapePlans(gates);
    cout << result << endl;

    return 0;
}

这段代码首先读取了输入的矩阵大小，然后读取了矩阵中每个单元格的解锁成本。最后，将调用 `countEscapePlans` 函数计算最小成本逃生计划的数量，并输出结果。

希望这可以帮助到你！如果你有任何其他问题，请随时问我。