HDOJ 题目2874 Connections between cities（LCA转RMQ+并查集）

最新推荐文章于 2020-05-04 15:32:54 发布

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最新推荐文章于 2020-05-04 15:32:54 发布

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分类专栏： LCA

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本文探讨了在战后城市重建中如何解决材料运输的问题，特别是在道路被破坏的情况下寻找两个城市间最短路径的方法。通过一种特殊的路径查找算法，文章详细介绍了如何在没有直接连接或者存在部分损坏道路的城市间找到最短路径。

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Connections between cities

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6998 Accepted Submission(s): 1812

Problem Description

After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.

Input

Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.

Output

For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.

Sample Input

Sample Output

  
   Not connected
6


   
    
     Hint
    
Hint

Huge input, scanf recommended.

Source

2009 Multi-University Training Contest 8 - Host by BJNU

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ac代码

#include<stdio.h>   
#include<string.h>   
#include<math.h>   
#include<algorithm>   
using namespace std;  
#define N 10010   
int vis[N];  
int first[N*2],node[N*2],deep[N*2],minv[N<<1][25],hah[N],pre[N],dis[N];  
int n,q,cnt,m;  
int head[N];  
struct s  
{  
    int u,v,w,next;  
}edge[N<<1];  
void add(int u,int v,int w)  
{  
    edge[cnt].u=u;  
    edge[cnt].v=v;  
	edge[cnt].w=w;
    edge[cnt].next=head[u];  
    head[u]=cnt++;  
}  
int tot;  
void dfs(int u,int dep)  
{  
    tot++;  
   // fa[u]=pre;  
    node[tot]=u;  
    deep[tot]=dep;  
    vis[u]=1;  
    first[u]=tot;  
    int i;  
    for(i=head[u];i!=-1;i=edge[i].next)  
    {  
        int v=edge[i].v;  
        if(!vis[v])  
        {  
			dis[v]=dis[u]+edge[i].w;
            dfs(v,dep+1);  
            tot++;  
            node[tot]=u;  
            deep[tot]=dep;  
        }  
    }  
}  
void init_RMQ(int n)    
{    
    int i,j,k;    
    for(i=1;i<=n;i++)    
    {    
        minv[i][0]=i;    
    }    
	int kk = (int) (log((double) n) / log(2.0));
    for(j=1;j<=kk;j++)    
    {    
        for(k=1;k+(1<<j)-1<=n;k++)    
        {    
            if(deep[minv[k][j-1]]>deep[minv[k+(1<<(j-1))][j-1]])  
                minv[k][j]=minv[k+(1<<(j-1))][j-1];  
            else  
                minv[k][j]=minv[k][j-1];  
        }    
    }    
}    
int q_min(int l,int r)    
{    
    int k=(int)(log((double)(r-l+1))/(log(2.0)));    
    if(deep[minv[l][k]]>deep[minv[r-(1<<k)+1][k]])  
        return minv[r-(1<<k)+1][k];  
    else  
        return minv[l][k];  
}   
int lca(int a,int b)  
{  
    int x=first[a];  
    int y=first[b];  
    int k;  
   if(x<y)
   {
	   k=q_min(x,y);
   }
   else
	   k=q_min(y,x);
   return node[k];  
}  
void init()
{
	int i;
	for(i=0;i<=n;i++)
	{
		pre[i]=i;
		node[i]=0;
	}
	memset(first,0,sizeof(first));
}
int find(int x)
{
	if(x==pre[x])
		return x;
	return pre[x]=find(pre[x]);
}
void merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
		pre[fx]=fy;
}
int main()
{
//	int n,m,q;
	while(scanf("%d%d%d",&n,&m,&q)!=EOF)
	{
		memset(head,-1,sizeof(head));
		cnt=0;
		int i,j;
		init();
		for(i=1;i<=m;i++)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			add(a,b,c);
			add(b,a,c);
			merge(a,b);
		}
		memset(hah,0,sizeof(hah));
		for(i=1;i<=n;i++)
		{
			if(pre[i]==i)
			{
				add(0,i,0);
				add(i,0,0);
			}
		}
		tot=0;
		memset(vis,0,sizeof(vis));
		dis[0]=0;
		dfs(0,0);
		init_RMQ(tot-1);
		while(q--)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			if(find(a)!=find(b))
			{
				printf("Not connected\n");
				continue;
			}
			int c=lca(a,b);
			printf("%d\n",dis[a]+dis[b]-2*dis[c]);
		}
	}
}