POJ-1789-Truck History
来源:互联网 发布:魔卡少女樱 知乎 编辑:程序博客网 时间:2024/05/06 22:58
Truck History
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 24666 Accepted: 9610
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
4aaaaaaabaaaaaaabaaaaaaabaaaa0
Sample Output
The highest possible quality is 1/3.
#include<string.h>#include<stdio.h>using namespace std;const int MAXN = 2001;const int INF = 0x3f3f3f;int Map[MAXN][MAXN], dis[MAXN];char s[MAXN][100];bool vis[MAXN];int f(char s1[], char s2[]) //判断两个字符串中同一位置不同字符的个数{ int k = 0; for(int i = 0;i<7;++i) { if(s1[i]!=s2[i])++k; } return k;}int main(){ int n; while(~scanf("%d",&n)&&n) { memset(Map,0,sizeof(Map)); for(int i= 0;i<n;++i) { scanf("%s",s[i]); for(int j = 0;j<i;++j) Map[i][j] = Map[j][i] = f(s[i], s[j]); //<span style="font-family: 'Courier New', Courier, monospace;"> 两点间距离就是字符串中不同字符的个数</span> } memset(vis,0,sizeof(vis)); int sum = 0, pos; for(int i = 0;i<n;++i) dis[i] = Map[0][i]; vis[0] = true; for(int i = 0; i<n-1;++i) //Prim(); { int MIN = INF; for(int j = 0;j<n;++j) { if(dis[j]<MIN&&!vis[j]) { MIN = dis[j]; pos = j; } } vis[pos] = true; sum+=MIN; for(int j = 0;j<n;++j) { if(!vis[j]&&Map[pos][j]<dis[j]) dis[j] = Map[pos][j]; } } printf("The highest possible quality is 1/%d.\n",sum); } return 0;}
0 0
- POJ 1789 Truck History
- POJ 1789 Truck History
- POJ 1789 Truck History
- poj 1789 Truck History
- poj 1789 Truck History
- Poj 1789 Truck History
- poj 1789 Truck History
- poj 1789 Truck History
- poj 1789 Truck History
- poj 1789 Truck History
- POJ-1789-Truck History
- poj-1789-Truck History
- POJ 1789 Truck History
- poj 1789 Truck History
- POJ 1789 Truck History
- POJ 1789 Truck History
- POJ 1789 Truck History
- POJ 1789 Truck History
- android--登录例子,保存用户名和密码到这个应用所在文件中,从文件中读取用户名和密码
- 洛谷1387最大正方形
- dtd 约束细则
- AFNnetworking详解
- POJ-2240-Arbitrage
- POJ-1789-Truck History
- 【Basic computer】-----Persist Object’s lifecycle in hibernate (Hibernate 中持久化对象的生命周期)
- 数据结构实验之队列一:排队买饭
- MySQL5.7在CentOS7下安装
- HDU 4044 GeoDefense 树形dp:树形背包★
- 【计算机网络】——网络层,IP地址,IP数据报,数据的分片及组装详解
- [JQ权威指南]JQuery读取XML数据
- Android Activity:四种启动模式,Intent Flags和任务栈
- [Cloud Computing]Patterns: Broad Access