poj 1789 Truck History 【prim & kruskal】

Truck History

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 21527 Accepted: 8374

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.嗯，题目大意就是说，给出卡车的编号，编号是一个长度为7的字符串，一个编号是由另一个编号衍生出来的，而衍生出来所需要的代价就是两个编号之间相同位置不同字符的个数（相当于两个顶点相连后的权值），现在求给出的这组字符串所需要的最小代价。

kruskal  1219ms#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,pre[2200];char s[2200][10];struct node{    int a,b,v;};node road[2200*2200];int cmp(node a,node b){    return a.v<b.v;}int find(int x){    if(x!=pre[x])        pre[x]=find(pre[x]);    return pre[x];}int check(int i,int j){    int ans=0;    for(int k=0;k<7;++k)        if(s[i][k]!=s[j][k])            ans++;    return ans;}int main(){    int fe;    while(scanf("%d",&n)&&n)    {        fe=0;        for(int i=1;i<=n;++i)            pre[i]=i;        for(int i=1;i<=n;++i)            scanf("%s",s[i]);        int k=0;        for(int i=1;i<=n;++i)        {            for(int j=1;j<i;++j)            {                if(i!=j)                {                    road[k].a=i;                    road[k].b=j;                    road[k].v=check(i,j);                    k++;                }            }        }        sort(road,road+k,cmp);        for(int i=0;i<k;++i)        {            int fa=find(road[i].a);            int fb=find(road[i].b);            if(fa!=fb)            {                pre[fa]=fb;                fe+=road[i].v;            }        }        printf("The highest possible quality is 1/%d.\n",fe);    }    return 0;}

 

//prim  610ms#include<cstdio>#include<cstring>#define inf 0x3f3f3f3fint n,map[2020][2020];int vis[2020],rec[2020];char s[2020][10];int check(int i,int j){    int ans=0;    for(int k=0;k<7;++k)        if(s[i][k]!=s[j][k])            ans++;    return ans;}void prim(){    int mark,minw,fe=0;    while(1)    {        mark=0;        minw=inf;        for(int i=1;i<=n;++i)        {            if(!vis[i]&&rec[i]<minw)            {                minw=rec[i];                mark=i;            }        }        if(!mark)            break;        fe+=minw;        vis[mark]=1;        for(int i=1;i<=n;++i)        {            if(!vis[i]&&rec[i]>map[mark][i])                rec[i]=map[mark][i];        }    }    printf("The highest possible quality is 1/%d.\n",fe);}int main(){    while(scanf("%d",&n)&&n)    {        memset(vis,0,sizeof(vis));        memset(rec,0,sizeof(rec));        memset(map,inf,sizeof(inf));        for(int i=1;i<=n;++i)            scanf("%s",s[i]);        for(int i=1;i<=n;++i)        {            for(int j=1;j<=n;++j)            {                map[i][j]=map[j][i]=check(i,j);            }        }        vis[1]=1;        for(int i=1;i<=n;++i)            rec[i]=map[1][i];        prim();    }    return 0;}