用C++实现以下全部问题

好的，以下是使用 C++ 实现斐波那契数列、Floyd 算法和 0/1 背包问题的代码示例：

### 1. 斐波那契数列

#include <iostream>
#include <vector>

int fibonacci(int n) {
    if (n <= 1) return n;
    
    std::vector<int> dp(n + 1, 0);
    dp[0] = 0;
    dp[1] = 1;
    
    for (int i = 2; i <= n; ++i) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    
    return dp[n];
}

int main() {
    int n;
    std::cout << "Enter the value of n: ";
    std::cin >> n;
    
    std::cout << "Fibonacci number at position " << n << " is " << fibonacci(n) << std::endl;
    
    return 0;
}

### 2. Floyd 算法

#include <iostream>
#include <vector>
#include <climits>

void floyd(std::vector<std::vector<int>>& graph, int V) {
    std::vector<std::vector<int>> dist(graph);

    // Initialize the distance matrix with the graph's edge weights
    for (int i = 0; i < V; ++i) {
        for (int j = 0; j < V; ++j) {
            if (graph[i][j] == 0 && i != j) {
                dist[i][j] = INT_MAX;
            }
        }
    }

    // Apply Floyd-Warshall algorithm
    for (int k = 0; k < V; ++k) {
        for (int i = 0; i < V; ++i) {
            for (int j = 0; j < V; ++j) {
                if (dist[i][k] != INT_MAX && dist[k][j] != INT_MAX && dist[i][k] + dist[k][j] < dist[i][j]) {
                    dist[i][j] = dist[i][k] + dist[k][j];
                }
            }
        }
    }

    // Print the shortest path matrix
    for (int i = 0; i < V; ++i) {
        for (int j = 0; j < V; ++j) {
            if (dist[i][j] == INT_MAX) {
                std::cout << "INF ";
            } else {
                std::cout << dist[i][j] << " ";
            }
        }
        std::cout << std::endl;
    }
}

int main() {
    int V;
    std::cout << "Enter the number of vertices: ";
    std::cin >> V;

    std::vector<std::vector<int>> graph(V, std::vector<int>(V, 0));

    std::cout << "Enter the adjacency matrix:\n";
    for (int i = 0; i < V; ++i) {
        for (int j = 0; j < V; ++j) {
            std::cin >> graph[i][j];
        }
    }

    floyd(graph, V);

    return 0;
}

### 3. 0/1 背包问题

#include <iostream>
#include <vector>

int knapsack(int W, const std::vector<int>& wt, const std::vector<int>& val, int n) {
    std::vector<std::vector<int>> dp(n + 1, std::vector<int>(W + 1, 0));
    
    for (int i = 1; i <= n; ++i) {
        for (int w = 1; w <= W; ++w) {
            if (wt[i - 1] + dp[i - 1][w - wt[i - 1]], dp[i - 1][w]);
            } else {
                dp[i][w] = dp[i - 1][w];
            }
        }
    }
    
    return dp[n][W];
}

int main() {
    int n, W;
    std::cout << "Enter the number of items: ";
    std::cin >> n;
    std::cout << "Enter the capacity of the knapsack: ";
    std::cin >> W;

    std::vector<int> val(n), wt(n);
    std::cout << "Enter the values of the items:\n";
    for (int i = 0; i < n; ++i) {
        std::cin >> val[i];
    }

    std::cout << "Enter the weights of the items:\n";
    for (int i = 0; i < n; ++i) {
        std::cin >> wt[i];
    }

    int max_value = knapsack(W, wt, val, n);
    std::cout << "The maximum value that can be obtained is: " << max_value << std::endl;

    return 0;
}

这些代码分别实现了斐波那契数列、Floyd 算法和 0/1 背包问题，并且都使用了动态规划的方法。希望这些代码对你有帮助！如果有任何问题或需要进一步解释，请随时告诉我。