文章目录
  1. 1. 原题重现
    1. 1.1. Description
    2. 1.2. Input
    3. 1.3. Output
    4. 1.4. Sample Input
    5. 1.5. Sample Output
  2. 2. 简单说说
  3. 3. 代码

引言:这道题是很基础的最小生成树问题,我一开始在题意理解上出现了很大的问题,从而用了最短路去做,浪费很多时间。


原题重现

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 t o is the original type and t d the type derived from it and d(t o,t d) 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

	4
	aaaaaaa
	baaaaaa
	abaaaaa
	aabaaaa
	0

Sample Output

	The highest possible quality is 1/3.

简单说说

这道题不难,如果理解了题意就知道这是一个最小生成树问题。
最小生成树就是说在N个点中找N-1条边,使得这N-1条边之和最短,用铺路的话就是说铺路代价最小,对应的,最短路问题是指两点之间最短的长度。

  • 关键语句1“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).”
    就是说两个卡车字符串之间的不同字符的数量定义为两个卡车的距离,现在给出一系列卡车的字符串,除了第一个卡车,其他卡车都是由第一个卡车演变而来的。注意,这里第一个不是指输入中的第一个。

  • 关键语句2“where the sum goes over all pairs of types in the derivation plan such that t o is the original type and t d the type derived from it and d(t o,t d) is the distance of the types. ”
    就是说看公式,d(to,td)代表两个卡车的距离,求演变出所有卡车所需距离的总和。
    我题意就在这个(to,td)混淆了,其实to代表的是origin而不是t0t_0

  • 理解后直接Kruskal算法贪心出N-1条边就行了,关键在于确定连通性,这个用并查集来确定。假如FIND(i)/=FIND(j)FIND(i)\rlap{\,/}{=} FIND(j)就说明两个点没连通。

代码

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#define _CRT_SECURE_NO_WARNINGS
#pragma comment (linker,"/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<ctime>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
#define MOD (ll)(1e9+7)
inline int read()
{
int a = 0, f = 1; char c = getchar();
while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
while (isdigit(c)) (a *= 10) += c - '0', c = getchar();
return a*f;
}
inline ll read2()
{
ll a = 0, f = 1; char c = getchar();
while (!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
while (isdigit(c)) (a *= 10) += c - '0', c = getchar();
return a*f;
}

int a[2005];
int N;
int Find(int n)
{
int i = n;
while (a[i] != i)
{
i = a[i];
}
int t;
while (a[n] != i)
{
t = a[n];
a[n] = i;
n = t;
}
return i;
}
void joint(int x, int y)
{
int tx = Find(x), ty = Find(y);
if (tx != ty)
a[tx] = ty;
}
struct node
{
int u, v, w;
};
vector<node> vec;
char str[2010][10];
bool vis[2010];
int e[2010][2010];
void dfs()
{
//puts("?");
for (int i = 0; i<N - 1; i++)
{
for (int j = i + 1; j<N; j++)
{
int w = 0;
for (int k = 0; k<7; k++)
if (str[i][k] != str[j][k])
w++;
//printf("|%d|", w);
vec.push_back(node{ i,j,w });
}
}
}
bool cmp(node a, node b)
{
return a.w<b.w;
}
int main(void)
{
while (1)
{
vec.clear();
N = read();
memset(e,INF,sizeof(e));
if (!N)
break;
for (int i = 0; i<N; i++)
{
a[i] = i;
scanf("%s", str[i]);
}
dfs();
sort(vec.begin(), vec.end(), cmp);
int cnt = 0, sum = 0;
for (int i = 0; cnt<N - 1; i++)
{
if (Find(vec[i].u) != Find(vec[i].v))
{
joint(vec[i].u, vec[i].v);
cnt++;
sum += vec[i].w;
}
}
printf("The highest possible quality is 1/%d.\n", sum);
}
return 0;
}
文章目录
  1. 1. 原题重现
    1. 1.1. Description
    2. 1.2. Input
    3. 1.3. Output
    4. 1.4. Sample Input
    5. 1.5. Sample Output
  2. 2. 简单说说
  3. 3. 代码