POJ1789 Truck History 最小生成树
2016-11-13 14:38
309 查看
Truck History
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.
Input
The input consists of several test cases. Each test case begins with a line containing the number of truck type
4000
s, 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
Sample Output
题目水就算了,还搞这么长。。。。看了四五次才大概的看明白了讲的是个什么东西,其实就是每两个truck type之间有一个距离,这个距离就是他们的字符串在相同位置上的不同的字母个数。。。然后就可以的到这个完全图的各边权值,然后就在里面找最小生成树就是了。
代码如下:
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxx=2005;
const int INF=1<<25;
string str[maxx];
int cost[maxx][maxx];
bool used[maxx];
int minroad[maxx];
int n;
void init(){
memset(used,0,sizeof(used));
for(int i=0;i<=n;i++)minroad[i]=INF;minroad[0]=0;
}
int prim(){
int res=0;
while(true)
{
int v=-1;
for(int i=0;i<n;i++){
if(!used[i]&&(v==-1||minroad[v]>minroad[i]))v=i;
}
if(v==-1)break;
used[v]=true;
res+=minroad[v];
for(int i=0;i<n;i++)
minroad[i]=min(minroad[i],cost[v][i]);
}
return res;
}
int main(){
while(cin>>n&&n){
init();
for(int i=0;i<n;i++)
cin>>str[i];
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j++)
{
if(i==j)cost[i][j]=0;
else {
int num=0;
for(int k=0;k<7;k++)if(str[i][k]!=str[j][k])num++;
cost[i][j]=num;
cost[j][i]=num;
}
}
}
int aa=prim();
cout<<"The highest possible quality is "<<"1/"<<aa<<"."<<endl;
}
return 0;}
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 25988 | Accepted: 10083 |
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.
Input
The input consists of several test cases. Each test case begins with a line containing the number of truck type
4000
s, 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 0
Sample Output
The highest possible quality is 1/3.
题目水就算了,还搞这么长。。。。看了四五次才大概的看明白了讲的是个什么东西,其实就是每两个truck type之间有一个距离,这个距离就是他们的字符串在相同位置上的不同的字母个数。。。然后就可以的到这个完全图的各边权值,然后就在里面找最小生成树就是了。
代码如下:
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxx=2005;
const int INF=1<<25;
string str[maxx];
int cost[maxx][maxx];
bool used[maxx];
int minroad[maxx];
int n;
void init(){
memset(used,0,sizeof(used));
for(int i=0;i<=n;i++)minroad[i]=INF;minroad[0]=0;
}
int prim(){
int res=0;
while(true)
{
int v=-1;
for(int i=0;i<n;i++){
if(!used[i]&&(v==-1||minroad[v]>minroad[i]))v=i;
}
if(v==-1)break;
used[v]=true;
res+=minroad[v];
for(int i=0;i<n;i++)
minroad[i]=min(minroad[i],cost[v][i]);
}
return res;
}
int main(){
while(cin>>n&&n){
init();
for(int i=0;i<n;i++)
cin>>str[i];
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j++)
{
if(i==j)cost[i][j]=0;
else {
int num=0;
for(int k=0;k<7;k++)if(str[i][k]!=str[j][k])num++;
cost[i][j]=num;
cost[j][i]=num;
}
}
}
int aa=prim();
cout<<"The highest possible quality is "<<"1/"<<aa<<"."<<endl;
}
return 0;}
相关文章推荐
- POJ1789 Truck History 最小生成树 + 路径压缩
- POJ 1789 Truck History 图论 prim算法 最小生成树
- POJ 1789-Truck History 最小生成树 Kruskal算法
- POJ-1789 Truck History 裸最小生成树
- 初级->图算法->最小生成树 poj 1789 Truck History
- poj 1789 Truck History 最小生成树
- POJ1789 Truck History(Prim最小生成树)
- poj 1789 Truck History 最小生成树
- poj1789 Truck History(最小生成树)
- poj1789 Truck History ——最小生成树入门题_Prim算法
- POJ1789 Truck History(最小生成树)
- POJ-1789 Truck History 最小生成树
- poj 1789 Truck History (最小生成树)
- POJ1789,Truck History,最小生成树,Prim
- POJ 1789 Truck History 最小生成树 KRUSKAL算法
- POJ 一 1789 Truck History(最小生成树)
- poj 1789 Truck History 最小生成树
- POJ 1789 Truck History【最小生成树简单应用】
- POJ1789--Truck History--最小生成树
- poj 1789 Truck History(最小生成树 prim)