11月27日 connect the city

最新推荐文章于 2024-04-01 21:54:41 发布

转载最新推荐文章于 2024-04-01 21:54:41 发布 · 177 阅读

算法题专栏收录该内容

35 篇文章

订阅专栏

本文探讨了一个关于通过构建道路来重新连接因海平面上升而断开的城市的问题。使用了图论中的并查集算法实现最小生成树，通过两种不同的代码实现方式解决了问题，并对比了性能。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

本题我一直TLE，所以代码转载自https://www.cnblogs.com/13224ACMer/p/4639351.html

Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

Sample Output

我自己的代码（感觉没问题，但是tle了，不知道是不是qsort的问题？）

#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
int rt[505], pr[505],Rank[505],ncase,sum;
struct edge{
	int a, b, le;
}eg[25005];
int n, m, k;
int cnt;
int find(int x)
{
	int tmp=x,root = x;
	while (pr[root] != root)
	{
		root = pr[root];
	}
	/*while (pr[tmp] != root)
	{
		rt[tmp] = root;
		tmp = pr[tmp];
	}*/
	return root;
}
void Union(int x,int y)
{
	int a = find(x);
	int b = find(y);
	pr[a] = b;
	/*else
	{
		pr[b] = a;
		if (Rank[a] == Rank[b])Rank[a]++;
	}*/
}
int cmp(const void* a, const void* b)
{
	return ((edge*)a)->le - ((edge*)b)->le;
}
void init(int n)
{
	for (int i = 0; i <= n; i++)
	{
		pr[i] = i;
		rt[i] = i;
	}
	sum = 0;
}
int main()
{
	scanf("%d", &ncase);
	for (int i = 0; i < ncase; i++)
	{
		int nedge = 0;
		scanf("%d%d%d", &n, &m, &k);
		int q = n;
		init(n);
		for (int j = 0; j < m; j++)
		{
			scanf("%d%d%d", &eg[j].a, &eg[j].b, &eg[j].le);
		}
		for (int j = 0; j < k; j++)
		{
			int num,rr;
			scanf("%d", &num);
			scanf("%d", &rr);
			for (int k = 0; k < num-1; k++)
			{
				int b;
				scanf("%d", &b);
				q -= (b-1);
				Union(b, rr);
			}
		}
		qsort(eg, m, sizeof(eg[0]), cmp);
		for (int j = 0; j < m && nedge!=q-1; j++)
		{
			int x = find(eg[j].a);
			int y = find(eg[j].b);
			if (x != y)
			{
				Union(eg[j].a, eg[j].b);
				sum += eg[j].le;
				nedge++;
			}
		}
		if (nedge < q - 1) printf("%d\n", -1);
		else printf("%d\n", sum);
	}
	return 0;
}

别人的代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
   int u;
   int v;
   int w;
}que[100000];
int father[505];
bool cmp(struct node a,struct node b){
    return a.w<b.w;
}
void init(int n){
    for(int i=1;i<=n;i++)
    father[i]=i;
}
int find(int u){
    if(father[u]!=u)
    father[u]=find(father[u]);
    return father[u];
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
       int n,m,k;
       int edge=0,sum=0;
       scanf("%d%d%d",&n,&m,&k);
       init(n);
       for(int i=0;i<m;i++){
          scanf("%d%d%d",&que[i].u,&que[i].v,&que[i].w);
       }
       sort(que,que+m,cmp);
       for(int i=1;i<=k;i++){
           int cnt,root;
           scanf("%d%d",&cnt,&root);
           for(int j=1;j<cnt;j++){
              int x;
              scanf("%d",&x);
              int c1=find(x),c2=find(root);
              if(c1!=c2){
                  father[c1]=c2;
              }
           }
       }
      for(int i=1;i<=n;i++)
      if(father[i]==i)
      edge++;
      bool flag=false;
      for(int i=0;i<m;i++){
          int tx=find(que[i].u),ty=find(que[i].v);
          if(tx!=ty){
              father[tx]=ty;
              sum+=que[i].w;
              edge--;
          }
          if(edge==1){
            flag=true;
            break;
          }
 
      }
      if(flag)
      printf("%d\n",sum);
      else
      printf("-1\n");
 
    }
    return 0;
}

希望大神指导指导