POJ 1789：Truck History（prim&&最小產生樹）

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Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17610		Accepted: 6786

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(to,td)
where the sum goes over all pairs of types in the derivation plan such that to is the original type and td the type derived from it and d(to,td) 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.

用一個7位的string代表一個編號，兩個編號之間的distance代表這兩個編號之間不同字母的個數。

一個編號只能由另一個編號“衍生”出來，代價是這兩個編號之間相應的distance，

現在要找出一個“衍生”方案，使得總代價最小，也就是distance之和最小。

此題的關鍵是將問題轉化為最小產生樹的問題。

每一個編號為圖的一個頂點，頂點與頂點間的編號差即為這條邊的權值，題目所要的就是我們求出最小產生樹來。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>#include<queue>#define INF 0x3f3f3fusing namespace std;const int maxn = 2000 + 50;int f[maxn];int map[maxn][maxn];bool vist[maxn];char str[maxn][10];int ans[maxn];int n;int find(int x, int y){    int cnt = 0;    for(int i=0; i<7; i++)        if( str[x][i]!=str[y][i] )        cnt++;    return cnt;}void init(){    memset( map, 0, sizeof(map) );    memset( vist, false, sizeof(vist) );    memset( ans, 0, sizeof(ans) );    for(int i=0; i<n; i++)        scanf( "%s", str[i] );    //for(int i=0; i<n; i++)    //    printf("%s\n", str[i]);    for(int i=0; i<n; i++)        for(int j=0; j<=i; j++)        map[i][j] = map[j][i] = find(i, j);}void prim(){    int minc, mind;    vist[0] = true;    ans[0] = 0;    for(int i=1; i<n; i++)        ans[i] = map[0][i];    for(int j=0; j<n-1; j++)    {        minc = INF;        for(int i=0; i<n; i++)        {            if( !vist[i] && minc>ans[i] )            {                minc = ans[i];                mind = i;            }        }       if(minc != INF)        {            vist[mind] = true;            for(int i=0; i<n; i++)                if( !vist[i] && ans[i]>map[mind][i] )                ans[i] = map[mind][i];        }    }}void output(){    int sum = 0;    for(int i=1; i<n; i++)        sum += ans[i];    printf("The highest possible quality is 1/%d.\n", sum);}int main(){    while( scanf( "%d", &n )==1 &&n  )    {        init();        prim();        output();    }    return 0;}