CCF CSP 数据中心 c++ python csp201812_4 100分

样例输入
4
5
1
1 2 3
1 3 4
1 4 5
2 3 8
3 4 2
样例输出
4
样例说明
　　下图是样例说明。

问题分析
        这道题理解了题意后，其实就是求图的最小生成树，那个根没有一点用，根本不用考虑，只需要求出最小生成树的最大权值就ok了。传统的求最小生成树的算法有Prim算法和kruskal算法。但这篇博客我使用并查集的方法去解决。如果对并查集不了解的可以先去查查并查集。具体代码如下：
c++代码如下：

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct Edge{
int from;
int to;
int distance;
};
int father[50005];
bool cmp(Edge e1,Edge e2){
return e1.distance < e2.distance;
}
int result = 0;
int main(){
void init(int number);
void unio(int from,int to,int distance);
int vertextNumber,edgeNumber,root;
scanf("%d%d%d",&vertextNumber,&edgeNumber,&root);
vector<Edge> vec;
Edge temp;
int from,to,distance;
for(int i = 0;i < edgeNumber;i++){
scanf("%d%d%d",&from,&to,&distance);
temp.from = from;
temp.to = to;
temp.distance = distance;
vec.push_back(temp);
}
sort(vec.begin(),vec.end(),cmp);
init(vertextNumber);
for(int i = 0;i < vec.size();i++){
temp = vec[i];
from = temp.from;
to = temp.to;
distance = temp.distance;
unio(from,to,distance);
}
printf("%d\n",result);
return 0;
}
void init(int number){
for(int i = 0;i <= number;i++){
father[i] = i;
}
}
int findFather(int x){
int a = x;
while(x != father[x]){
x = father[x];
}
while(a != father[a]){
int z = father[a];
father[a] = x;
a = z;
}
return x;
}
void unio(int from,int to,int distance){
int fx = findFather(from);
int fy = findFather(to);
if(fx != fy){
father[fx] = fy;
result = distance;
}
}

java代码如下：

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class csp201812_4 {
public static int edgeNumber;
public static int[] father;
public static int result;
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
edgeNumber = input.nextInt();
int number = input.nextInt();
int root = input.nextInt();
List<Edge> list = new ArrayList<>();
int from,to,distance;
for(int i = 0;i < number;i++) {
from = input.nextInt();
to = input.nextInt();
distance = input.nextInt();
list.add(new Edge(from,to,distance));
}
init();
Collections.sort(list);
Edge temp;
for(int i = 0;i < list.size();i++) {
temp = list.get(i);
Union(temp.from,temp.to,temp.distance);
}
System.out.println(result);
input.close();
}
public static void init() {
result = 0;
father = new int[edgeNumber + 1];
for(int i = 0;i < father.length;i++) {
father[i] = i;
}
}
public static int findFather(int x) {
int a = x;
while(x != father[x]) {
x = father[x];
}
while(a != father[a]) {
int z = a;
a = father[a];
father[z] = x;
}
return x;
}
public static void Union(int x,int y,int dd) {
int fx = findFather(x);
int fy = findFather(y);
if(fx != fy) {
father[fx] = fy;
result = dd;
}
}
}
class Edge implements Comparable<Edge>{
public int from;
public int to;
public int distance;
public Edge(int from, int to, int distance) {
super();
this.from = from;
this.to = to;
this.distance = distance;
}
@Override
public int compareTo(Edge o) {
return this.distance > o.distance ? 1 : this.distance < o.distance ? -1 : 0;
}
}

由于java代码运行时间比较长，故提交后运行超时，只能得80分。
python3代码如下：

from operator import attrgetter
class Edge(object):
def __init__(self,From,to,distance):
self.From = From
self.to = to
self.distance = distance
father = []
result = 0
def init(number):
for i in range(number):
father.append(i)
def findFather(x):
a = x
while x != father[x]:
x = father[x]
while a != father[a]:
z = father[a]
father[a] = x
a = z
return x
def union(From, to, distance):
global result
fx = findFather(From)
fy = findFather(to)
if fx != fy:
father[fx] = fy
result = distance
if __name__ == '__main__':
vertexNumber = (int)(input())
edgeNumber = (int)(input())
root = (int)(input())
lists = []
for i in range(edgeNumber):
temp = input().split(" ")
From = (int)(temp[0])
to = (int)(temp[1])
distance = (int)(temp[2])
edge = Edge(From, to, distance)
lists.append(edge)
init(vertexNumber + 1)
lists = sorted(lists, key=attrgetter('distance'))
for i in range(edgeNumber):
temp = lists[i]
From = temp.From
to = temp.to
distance = temp.distance
union(From, to, distance)
print(result)

ok!大功告成了，如果你有更好的方法，可以在评论区交流哦！