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HDU 2874 LCA离线算法

2024-07-15 14:01:00   212人阅读

Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4477    Accepted Submission(s): 1284


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
5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5
 

 

Sample Output
Not connected
6
 
 
 
题意：
第一行三个数n, m, q即为n个点，m条边，q次询问。下面m行为x, y, z即编号为x和y点之间有边权值为z，下面q次询问，每行x, y即询问x,y两点最短距离。
若x, y之间有边输出最短距离，否则输出Not connected。
 
思路：
第一次看到这个题想到是最短路径类型的题，提交上wa了，仔细看了一下题，发现给出的图是无环图也就是树或者森林，那么这个题很容易就知道是lca了。
假设x, y两点在一棵树上，它们的最近公共祖先为z。设x, y所在树的根结点为w，dis[x]即为点x到其树根的距离，那么x, y之间最短距离很明显就是
dis[x]+dis[y]-2*dis[z]。 那么用tarjan算法深搜一下就行了。
 
 
代码：
 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 #include <iostream> 6 using namespace std; 7 #define N 10005 8  9 struct node{10     int y, w;11 };12 13 int f[N];14 int father[N];15 int dis[N];16 int visited[N];17 int ans[1000005];            //注意ans数组的大小，我就RE了很长时间。。。 18 vector<node>ve[N], q[N];19 int n;20 21 void init(){22     for(int i=0;i<=n;i++){23         ve[i].clear();24         q[i].clear();25         f[i]=i;26         father[i]=i;27     }28     memset(dis,0,sizeof(dis));29     memset(visited,0,sizeof(visited));30     memset(ans,-1,sizeof(ans));31 }32 33 int findroot(int p){34     int r=p;35     while(r!=father[r])36     r=father[r];37     return r;38 }39 void lca(int u,int d,int ff){40     int v, i;41     node p;42     visited[u]=1;dis[u]=d;f[u]=ff;43     for(i=0;i<ve[u].size();i++){44         p=ve[u][i];45         v=p.y;46         if(!visited[v]){47             lca(v,d+p.w,ff);48             father[v]=u;49         }50     }51     for(i=0;i<q[u].size();i++){52         p=q[u][i];53         int id=p.w;54         v=p.y;55         if(visited[v]&&f[v]==f[u]){56             ans[id]=dis[u]+dis[v]-2*dis[findroot(v)];     //findroot(v)而不是findroot(u)，想一想为什么 57         }58     }59 }60 main()61 {62     int m, Q;63     int x, y, z;64     int i, j, k;65     node p;66     while(scanf("%d %d %d",&n,&m,&Q)==3){67         init();68         for(i=0;i<m;i++){69             scanf("%d %d %d",&x,&y,&z);70             p.y=y;p.w=z;71             ve[x].push_back(p);72             p.y=x;73             ve[y].push_back(p);74         }75         for(i=1;i<=Q;i++){76             scanf("%d %d",&x,&y);77             p.y=y;p.w=i;78             q[x].push_back(p);79             p.y=x;80             q[y].push_back(p);81         }82         for(i=1;i<=n;i++){83             if(!visited[i])84             lca(i,0,i);85         }86         for(i=1;i<=Q;i++){87             if(ans[i]==-1) printf("Not connected\n");88             else printf("%d\n",ans[i]);89         }90     }91 }

 


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