阿里在线编程测试

虽然编程测试没在规定时间内做出来，而且素质测试也凉透了，但是还是放一下后来打的代码，不知道能过多少样例。
问题描述
第一行输入两个数字，第一个代表系统个数，第二行代表系统之间的依赖关系。
然后输入每个系统的功耗，用空格隔开；最后输入各个系统之间的依赖关系。
需要计算出最大的耗时总量。（不记得是最长链还是最大耗时总量）

比如：

5 4
7 //1号系统耗时7ms
5 //2号系统耗时5ms
3 // ...
5
6
1 2 //1号系统依赖于2号系统
1 3 //1号系统依赖于3号系统
2 5 //2号系统依赖于5号系统
4 5 //4号系统依赖于5号系统
则依赖链有1->2->5,  1->3,  4->5

#include
using namespace std;
class Solution {
public:
    //深度优先算法求每条路径的总耗时，记录最大总耗时
    void dfs(int **matrix, bool* visited, int* cost, int point, int count, int& costs, int& ans) {
        if (!visited[point]) {
            visited[point] = true;
            for (int i = 1; i <= count; i++) {
                if (i != point && !visited[i] && matrix[point][i]) {
                    costs += cost[i];
                    //记录最大耗时总量
                    if (ans < costs) ans = costs;
                    dfs(matrix, visited, cost, i, count, costs, ans);
                    costs -= cost[i];
                }
            }
        }
    }
    void solution() {
        int count_system, count_dep;
        //输入系统数和依赖数
        cin >> count_system >> count_dep;

        //记录每个系统的耗时
        int* cost = new int[count_system+1];

        //记录依赖关系
        int** matrix = new int*[count_system+1];
        bool* visited = new bool[count_system + 1];

        for (int i = 0; i < count_system+1; i++) {
            matrix[i] = new int[count_system+1];
            for (int j = 0; j < count_system+1; j++) {
                matrix[i][j] = 0;
            }
        }
        //输入耗时和依赖关系
        for (int i = 1; i < count_system+1; i++) {
            cin >> cost[i];
        }
        for (int i = 0; i < count_dep; i++) {
            int start, end;
            cin >> start >> end;
            matrix[start][end] = 1;
        }
        int ans = 0;
        for (int i = 0; i < count_system; i++) {
            for (int i = 0; i < count_system + 1; i++) {
                visited[i] = false;
            }
            dfs(matrix, visited, cost, i + 1, count_system, cost[i+1], ans);
        }
        cout << ans << endl;

    }
};

int main() {
    Solution s;
    s.solution();
    system("pause");
}

/**若求最长链的总耗时
**则算法如下
**原理与上面的相似
**/

class Solution2 {
public:
    void dfs(int **matrix, bool* visited, int* cost, int point, int count, int length, int& longest, int costs, int& ans) {
        if (!visited[point]) {
            visited[point] = true;
            for (int i = 1; i <= count; i++) {
                if (i != point && !visited[i] && matrix[point][i]) {
                    costs += cost[i];
                    length += 1;
                    //记录最长链的耗时总量
                    if (longest < length) {
                        longest = length;
                        ans = costs;
                    }
                    dfs(matrix, visited, cost, i, count, length, longest, costs, ans);
                    costs -= cost[i];
                }
            }
        }
    }
    void solution() {
        int count_system, count_dep;
        cin >> count_system >> count_dep;
        int* cost = new int[count_system+1];
        int** matrix = new int*[count_system+1];
        bool* visited = new bool[count_system + 1];
        for (int i = 0; i < count_system+1; i++) {
            matrix[i] = new int[count_system+1];
            visited[i] = false;
            for (int j = 0; j < count_system+1; j++) {
                matrix[i][j] = 0;
            }
        }

        for (int i = 1; i < count_system+1; i++) {
            cin >> cost[i];
        }
        for (int i = 0; i < count_dep; i++) {
            int start, end;
            cin >> start >> end;
            matrix[start][end] = 1;
        }
        int ans = 0;
        int length = 0;
        for (int i = 0; i < count_system; i++) {
            for (int i = 0; i < count_system + 1; i++) {
                visited[i] = false;
            }
            dfs(matrix, visited, cost, i + 1, count_system, 0, length, cost[i + 1], ans);
        }
        cout << ans << endl;

    }
};