http://poj.org/problem?id=1789
Description
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
Output
Sample Input
4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0
Sample Output
The highest possible quality is 1/3.
此题难点在于读题,明白题意后就是一个很简单的最小生成树
明白一点就行:将每一个卡车类型代码(truck type codes)作为一个结点,任意两个 卡车类型代码中 相同位置即字符数组a[i],a[i+1] 上 a[i]和a[i+1]为不同字符的位置的个数 做为
这两个结点之间的路径的权值。
读题啊,硬伤啊!
#include <iostream>
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
using namespace std;
char a[][];
int map[][];
int n,dis[],v[];
void prim()
{
int min,sum=,k;
for(int i=; i<=n; i++)
{
v[i]=;
dis[i]=INF;
}
for(int i=; i<=n; i++)
dis[i]=map[][i];
v[]=;
for(int j=; j<n; j++)
{
min=INF;
for(int i=; i<=n; i++)
{
if(v[i]==&&dis[i]<min)
{
k=i;
min=dis[i];
}
}
sum+=min;
v[k]=;
for(int i=; i<=n; i++)
{
if(v[i]==&&map[k][i]<dis[i])
{
dis[i]=map[k][i];
}
}
}
cout<<"The highest possible quality is 1/"<<sum<<"."<<endl;
}
int main()
{
int count;
while(scanf("%d",&n)!=EOF&&n!=)
{
for(int i=; i<=n; i++)
scanf("%*c%s",a[i]);
for(int i=; i<=n; i++)
{
for(int j=i+; j<=n; j++)
{
count=;
for(int k=; k<; k++)
if(a[i][k]!=a[j][k])
{
count++;
}
map[i][j]=count;
map[j][i]=count;
}
}
prim();
}
return ;
}