POJ 1789 Truck History 最小生成树 prim

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 22429   Accepted: 8695 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 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 4 aaaaaaa baaaaaa abaaaaa aabaaaa 0 Sample Output The highest possible quality is 1/3. Source CTU Open 2003

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题意大概是这样的有n辆卡车每辆卡车的车牌号由7个字母组成，每一个车牌都可以由另一个车牌号转化成，代价是改变的数字。

找到改变最小字母的代价使得由一个车牌转化出所有的车牌号。

典型的最小生成树问题。

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 2005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
using namespace std;
bool vis[maxn];
int lowc[maxn];
int cost[maxn][maxn];
string str[maxn];
int get_v(int x,int y){
    int v=0;
    FOR(i,0,6)
        if(str[x][i]!=str[y][i])
            v++;
    return v;
}
int prim(int n){
    int ans=0;
    MT(vis,false);
    vis[0]=true;
    FOR(i,1,n)lowc[i]=cost[0][i];
    FOR(i,1,n-1){
        int minc=INF;
        int p=-1;
        FOR(j,0,n-1)
            if(!vis[j]&&minc>lowc[j]){
                minc=lowc[j];
                p=j;
            }
        if(minc==INF)return -1;
        ans+=minc;
        vis[p]=true;
        FOR(j,0,n-1)
            if(!vis[j]&&lowc[j]>cost[p][j])
                lowc[j]=cost[p][j];
    }
    //cout<<ans<<'\12';
    return ans;
}
int main(){
    int n;
    while(rd(n)&&n){
        FOR(i,0,n-1)
            cin>>str[i];
        FOR(i,0,n-1)
            FOR(j,0,n-1)
               cost[i][j]= get_v(i,j);
        printf("The highest possible quality is 1/%d.\n",prim(n));
    }
    return 0;
}
/*
4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0
*/