POJ 1789 Truck History

Truck History

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 23843 Accepted: 9251

Description

Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of exactly seven lowercase letters (each letter on each position has a very special meaning but that is unimportant for this task). At the beginning of company's history, just a single truck type was used but later other types were derived from it, then from the new types another types were derived, and so on. 

Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan -- i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as 
1/Σ_(to,td)d(t_o,t_d)
where the sum goes over all pairs of types in the derivation plan such that t_o is the original type and t_d the type derived from it and d(t_o,t_d) is the distance of the types. 
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan. 

Input

The input consists of several test cases. Each test case begins with a line containing the number of truck types, N, 2 <= N <= 2 000. Each of the following N lines of input contains one truck type code (a string of seven lowercase letters). You may assume that the codes uniquely describe the trucks, i.e., no two of these N lines are the same. The input is terminated with zero at the place of number of truck types.

Output

For each test case, your program should output the text "The highest possible quality is 1/Q.", where 1/Q is the quality of the best derivation plan.

Sample Input

4aaaaaaabaaaaaaabaaaaaaabaaaa0

Sample Output

The highest possible quality is 1/3.

【分析】

          此题为最小生成树问题，用领接矩阵记录两车之间的距离，距离是对应行数之间不同的字符数，然后采用Prim算法进行求解。

【代码】

<span style="font-size:18px;">#include <iostream>#include <cstdio>#include <queue>using namespace std;#define maxn 2020#define INF 10    //预定于的最大值int n;   //顶点数、边数int g[maxn][maxn];      //邻接矩阵表示struct node     //保存key值的结点{    int v;    int key;    friend bool operator<(node a, node b)   //自定义优先级，key小的优先    {        return a.key > b.key;    }};bool visited[maxn]; //是否已经加入树种node vx[maxn];      //保存每个结点与其父节点连接边的权值priority_queue<node> q; //优先队列stl实现/* 用优先队列来做省去了比较出最小值，节约时间，但是代码变复杂了很多*/void Prim(int s)    //s表示根结点{    for(int i = 1; i <= n; i++) //初始化    {        vx[i].v = i;        vx[i].key = INF;        visited[i] = false;  // 表示没遍历过    }    vx[s].key = 0;    q.push(vx[s]);    while(!q.empty())    {        node nd = q.top();  //取队首，记得赶紧pop掉        q.pop();        if(visited[nd.v] == true)   // 遍历过就跳出，用来去除优先队列除队首外的元素            continue;        int st = nd.v;        visited[nd.v] = true;        for(int j = 1;  j <= n; j++)        {            /*              1.   j!=st表示不是自己，              2.   !visited[j]表示只改变没有被访问的元素，访问过的元素已经是最短，再进行变化可能会产生误差，              3.   最后一个表示与st的距离如果小于之前求的最小值就插入到优先队列里面，等除本身外的元素全插入完，最小的在                第一个，先执行最小的，可能最小的不一定总长度最长所以都遍历了一遍，但优先数列第一步最优解，剩下的也保存，                但是这个点已经走过，被标记为true，直接continue跳出            */            if(j!=st && !visited[j] && g[st][j] < vx[j].key)            {                vx[j].key = g[st][j];                q.push(vx[j]);            }        }    }}char tk[2020][8];int main(){    while(scanf("%d", &n), n)    {        for(int i = 1; i <= n; i++)            scanf("%s", tk[i]);        for(int i = 1; i < n; i++)      // 第i行        {            for(int j = i+1; j <= n; j++)   // 第j行            {                int cnt = 0;                for(int k = 0; k < 7; k++)  // 第k列                    if(tk[i][k] != tk[j][k])    // 第i行的元素与第j行的元素进行比较                        cnt++;                g[i][j] = g[j][i] = cnt;    // 一行中字符串与另一行字符串字符不一样的个数就是两车之间的距离            }        }        Prim(1);        int ans = 0;        for(int i = 1; i <= n; i++)            ans += vx[i].key;        printf("The highest possible quality is 1/%d.\n", ans);    }    return 0;}</span>

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