我只能說,題目描述的好像蠻厲害的樣子~
其實還只是一個prim的模板題~Orz
題意:給你N個字符串,每個字符串都只有7位,每兩個字符串會有一個“距離”,所謂的距離就是從0號位---6號位,一一對應過去,有1個字符不一樣就+1.所以距離就是兩個字符串的距離就是【0,7】,這樣就可以把每兩個字符串的距離算出來,而且第一個字符串不是派生出來的,所以就可以從第一個走起,建一個最小生成樹。
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 14101 | Accepted: 5366 |
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
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.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
const int M = 2005;
int n;
char name[M][10];
bool visit[M];
int mp[M][M];
int dis[M];
void prim( int st )
{
for( int i=0 ; i<n ; i++ ){
visit[i] = false;
dis[i] = mp[st][i];
}
dis[st] = 0;
visit[st] = true;
int ans = 0;
int tmp , k;
for( int i=0 ; i<n ; i++ ){
tmp = 99999999;
for( int j=0 ; j<n ; j++ ){
if( !visit[j] && tmp > dis[j] )
tmp = dis[ k = j ];
}
if( tmp == 99999999 )
break;
ans += tmp;
visit[k] = true;
for( int j=0 ; j<n ; j++ ){
if( !visit[j] && dis[j] > mp[k][j] )
dis[j] = mp[k][j];
}
}
printf("The highest possible quality is 1/%d.\n",ans);
}
int main()
{
while( scanf("%d",&n) == 1 && n ){
for( int i=0 ; i<n ; i++ )
scanf("%s",name[i]);
for( int i=0 ; i<n ; i++ ){
for( int j=i+1 ; j<n ; j++ ){
mp[i][j] = 0;
for( int k=0 ; k<7 ; k++ ){
if( name[i][k] != name[j][k] )
mp[i][j]++;
}
mp[j][i] = mp[i][j];
}
}
prim( 0 );
}
return 0;
}