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  • poj1789(prim)

    prim和kruskal都是求解最小生成树的算法。这道题题意就是有N个字符串就是N个节点,而字符串之间的距离就是节点边的长度,求其最小生成树的边权和。

    由于是第一次用prim,所以在求安全边的时候采用的是暴力的方法,所以我这个算法是O(n^2)的,跑了近1500ms,吓出一身冷汗……如果采用优先队列或者堆等数据结构应该会快,但是代码也会相应复杂了。

    #include <iostream>
    #include <string>
    using namespace std;
    
    int n;
    string str[2001];
    int dis[2001][2001];
    int key[2001];
    bool visited[2001];
    
    const int LEN = 7;
    
    int calDis(string str1, string str2)
    {
        int ans = 0;
        for (int i = 0; i < LEN; i++){
            if (str1[i] != str2[i]){
                ans++;
            }
        }
        return ans;
    }
    
    int prime()
    {
        int sum = 0;
        visited[0] = true;
        for (int i = 0; i < n; i++){
            key[i] = dis[0][i];
        }
        for (int i = 1; i < n; i++){
            int min = 100;
            int index = 0;
            for (int j = 0; j < n; j++){
                if (!visited[j] && key[j] < min){
                    index = j;
                    min = key[j];
                }
            }
            visited[index] = true;
            sum += key[index];
            for (int j = 0; j < n; j++){
                if (!visited[j] && dis[index][j] < key[j]){
                    key[j] = dis[index][j];
                }
            }
        }
        return sum;
    }
    
    int main()
    {
        while (cin >> n && n){
            memset(dis, 0, sizeof(dis));
            memset(key, 0, sizeof(key));
            memset(visited, 0, sizeof(visited));
            for (int i = 0; i < n; i++){
                cin >> str[i];
            }
            for (int i = 0; i < n; i++){
                for (int j = i; j < n; j++){
                    int tmpDis = calDis(str[i], str[j]);
                    dis[i][j] = tmpDis;
                    dis[j][i] = tmpDis;
                }
            }
            cout << "The highest possible quality is 1/" << prime() << '.' << endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/caiminfeng/p/4915365.html
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