Truck History
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 25609 | Accepted: 9979 |
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.
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
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
4 aaaaaaa baaaaaa abaaaaa aabaaaa
Sample Output
The highest possible quality is 1/3.
Source
题解：
题目大意是给你多个长度为7的字符串，然后每两个字符串之间的距离定义为这两个串对应位置上不同字符的个数
例如：aaaaaaa与abaaaaa对应位置上不同字符只有一个，所以距离为1。
然后问你从任意一个串出发，最少多少次变换才能得到所有给出的串，新变换出来的串也能变换成别的串。
所以，很明显，每两个串可以看成两个点，那么这就形成了一个完全图，那么问题就成了遍历整张图的最小价值，从而转化成了最小生成树算法。
注意，这道题是稠密图，所以要用prim算法。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char a[23333][10];
int d[233333];
int n;
bool used[23333];
int map[2005][2005];
int prim()
{
memset(used, 0, sizeof(used));
memset(d, 0x3f, sizeof(d));
d[1] = 0;
int res = 0;
while(1)
{
int v = -1;
for(int u = 1;u <= n;u ++)
{
if(!used[u] && (v == -1|| d[u] < d[v]))
v = u;
}
if(v == -1)
break;
used[v] = 1;
res += d[v];
for(int u = 1;u <= n;u ++)
{
d[u] = min(d[u], map[v][u]);
}
}
return res;
}
int main()
{
while(scanf("%d", &n)!=EOF)
{
if(n == 0)
break;
for(int i = 1;i <= n;i ++)
scanf("%s", a[i]+1);
int ans = 0;
for(int i = 1;i <= n;i ++)
{
for(int k = i;k <= n;k ++)
{
ans = 0;
if(k == i)
continue;
for(int j = 1;j <= 7;j ++)
{
if(a[i][j] != a[k][j])
{
ans ++;
}
}
map[i][k] = ans;
map[k][i] = ans;
}
}
printf("The highest possible quality is 1/%d.\n", prim());
}
return 0;
}