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发布时间:2022-07-27 13:56:01
 
Truck History

Time Limit: 5 Seconds      Memory Limit: 32768 KB

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 

 

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 Specification

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 Specification

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.

 

 

 

 

 

 

Sample Output

The highest possible quality is 1/3.

#include <cstdio> #include <cstdlib> #include <cstring> #define INF 1000000 #define MAXN 2000 //卡车类型数目的最大值 #define CODELEN 7 //编码长度 int N; //卡车类型数目 char codes[MAXN][CODELEN+3]; //存储每种卡车类型编码 int d[MAXN][MAXN]; //每对卡车类型之间的距离(邻接矩阵) int lowcost[MAXN]; //充当prim算法中的2个数组(lowcost数组和nearvex数组)的作用 int min_tree( ) { int i, j, k; //循环变量 int dist; //两个类型编码之间的距离 memset( d, 0, sizeof(d) ); for( i = 0; i < N; i++ ) //求第i种类型与第j种类型编码之间的距离 { for( j = i+1; j < N; j++ ) { dist = 0; for( k = 0; k < 7; k++ ) dist += codes[i][k] != codes[j][k]; d[i][j] = d[j][i] = dist; } } int sum = 0;//sum为最终求得的最小生成树的权值 lowcost[0] = -1; //从顶点0开始构造最小生成树 for( i = 1; i < N; i ++ ) lowcost[i] = d[0][i]; for( i = 1; i < N; i++ ) //把其他N-1个顶点扩展到生成树当中 { int min = INF; for( k = 0; k < N; k++ ) //找到当前可用的权值最小的边 { if( lowcost[k] != -1 && lowcost[k] < min ) { j = k; min = lowcost[k]; } } sum += min; lowcost[j] = -1; //把顶点j扩展进来 for( k = 0; k < N; k ++ ) { if( d[j][k] < lowcost[k] ) lowcost[k] = d[j][k]; } } return sum; } int main( ) { int i; //循环变量 while( 1 ) { scanf( "%d", &N ); if( N==0 ) break; for( i = 0; i < N; i++ ) scanf( "%s", codes[i] ); printf( "The highest possible quality is 1/%d. ", min_tree( ) ); } return 0; }

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