hdu 2874 Connections between cities (并查集+LCA)
Connections between cities
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4057 Accepted Submission(s): 1178
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.
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
Hint
Hint
Huge input, scanf recommended.
Source
Recommend
题意：
给出n个点m条边的的森林，求两个点间的最短距离。
并查集+LCA：
这题想了挺久的，觉得是比较经典的题目。首先要知道这是一个森林，和一般的LCA不同，要变成LCA来写其实也不难(没想方法，最后看别人思路才恍然大悟)，就是在所有树的根节点连上一个虚拟根节点0，将森林变成树，然后之前的节点用并查集处理，用于判断是否处于同棵树，处于同一个树再用LCA解决距离问题。
在线算法：
1 //2328MS 3728K 2757 B G++ 2 #include<stdio.h> 3 #include<string.h> 4 #include<math.h> 5 #define N 20005 6 struct node{ 7 int u,v,d; 8 int next; 9 }edge[2*N]; 10 int num,n,head[N]; 11 int root[2*N],rank[N]; 12 int dep[2*N],dis[N]; 13 int dp[2*N][30],vis[N]; 14 int set[N]; 15 void addedge(int u,int v,int d) 16 { 17 edge[num].u=u; 18 edge[num].v=v; 19 edge[num].d=d; 20 edge[num].next=head[u]; 21 head[u]=num++; 22 } 23 int Min(int a,int b) 24 { 25 return dep[a]<dep[b]?a:b; 26 } 27 int find(int x) 28 { 29 if(set[x]!=x) 30 set[x]=find(set[x]); 31 return set[x]; 32 } 33 void merge(int a,int b) 34 { 35 int x=find(a); 36 int y=find(b); 37 if(x>y) set[x]=y; 38 else set[y]=x; 39 } 40 void dfs(int u,int deep) 41 { 42 vis[u]=1; 43 root[++num]=u; 44 rank[u]=num; 45 dep[num]=deep; 46 for(int i=head[u];i!=-1;i=edge[i].next){ 47 int v=edge[i].v,d=edge[i].d; 48 if(vis[v]) continue; 49 dis[v]=dis[u]+d; 50 dfs(v,deep+1); 51 root[++num]=u; 52 dep[num]=deep; 53 } 54 } 55 void init() 56 { 57 int nn=2*n-1; 58 int m=(int)(log(nn*1.0)/log(2.0)); 59 for(int i=1;i<=nn;i++) 60 dp[i][0]=i; 61 for(int j=1;j<=m;j++) 62 for(int i=1;i+(1<<j)-1<=nn;i++) 63 dp[i][j]=Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); 64 } 65 int RMQ(int l,int r) 66 { 67 int m=(int)(log((r-l+1)*1.0)/log(2.0)); 68 return Min(dp[l][m],dp[r-(1<<m)+1][m]); 69 } 70 int LCA(int u,int v) 71 { 72 int a=rank[u]; 73 int b=rank[v]; 74 if(a>b) return root[RMQ(b,a)]; 75 else return root[RMQ(a,b)]; 76 } 77 int main(void) 78 { 79 int m,c,u,v,d; 80 int in[N]; 81 while(scanf("%d%d%d",&n,&m,&c)!=EOF) 82 { 83 memset(edge,0,sizeof(edge)); 84 memset(vis,0,sizeof(vis)); 85 memset(head,-1,sizeof(head)); 86 memset(in,0,sizeof(in)); 87 for(int i=0;i<=n;i++) set[i]=i; 88 num=0; 89 for(int i=0;i<m;i++){ 90 scanf("%d%d%d",&u,&v,&d); 91 addedge(u,v,d); 92 addedge(v,u,d); 93 in[v]++; 94 merge(u,v); 95 } 96 for(int i=1;i<=n;i++) //连接森林的所有树 97 if(in[i]==0){ 98 addedge(0,i,0); 99 addedge(i,0,0); 100 } 101 num=0; 102 dis[0]=0; 103 dfs(0,1); 104 init(); 105 //for(int i=1;i<=2*n-1;i++) printf(i==2*n-1?"%d\n":"%d ",root[i]); 106 //for(int i=1;i<=2*n-1;i++) printf(i==2*n-1?"%d\n":"%d ",dep[i]); 107 //for(int i=1;i<=2*n-1;i++) printf(i==2*n-1?"%d\n":"%d ",dis[i]); 108 //for(int i=1;i<=2*n-1;i++) printf(i==2*n-1?"%d\n":"%d ",rank[i]); 109 while(c--){ 110 scanf("%d%d",&u,&v); 111 //printf("*%d %d\n",find(u),find(v)); 112 if(find(u)!=find(v)) puts("Not connected"); 113 else printf("%d\n",dis[u]+dis[v]-2*dis[LCA(u,v)]); 114 } 115 } 116 return 0; 117 }
Tarjan离线做法：
1 //2359MS 30504K 1830B G++ 2 #include<iostream> 3 #include<queue> 4 #include<vector> 5 #define N 10005 6 using namespace std; 7 struct node{ 8 int v,d; 9 node(int a,int b){ 10 v=a;d=b; 11 } 12 }; 13 vector<node>child[N],V[N]; 14 int fa[N],vis[N],dis[N],mark[N]; 15 int res[100*N]; 16 int n; 17 int find(int x) 18 { 19 if(x!=fa[x]) fa[x]=find(fa[x]); 20 return fa[x]; 21 } 22 void Tarjan(int u) 23 { 24 fa[u]=u; 25 vis[u]=1; 26 for(int i=0;i<V[u].size();i++){ 27 int v=V[u][i].v,d=V[u][i].d; 28 if(vis[v] && res[d]==-1 && !mark[find(v)]) 29 res[d]=dis[u]+dis[v]-2*dis[find(v)]; 30 } 31 for(int i=0;i<child[u].size();i++){ 32 int v=child[u][i].v,d=child[u][i].d; 33 if(vis[v]) continue; 34 dis[v]=dis[u]+d; 35 Tarjan(v); 36 fa[v]=u; 37 } 38 } 39 int main(void) 40 { 41 int m,c; 42 int u,v,d; 43 while(scanf("%d%d%d",&n,&m,&c)!=EOF) 44 { 45 for(int i=0;i<=n;i++){ 46 child[i].clear(); 47 V[i].clear(); 48 fa[i]=i; 49 vis[i]=mark[i]=0; 50 } 51 memset(res,-1,sizeof(res)); 52 for(int i=1;i<=m;i++){ 53 scanf("%d%d%d",&u,&v,&d); 54 child[u].push_back(node(v,d)); 55 child[v].push_back(node(u,d)); 56 } 57 for(int i=1;i<=c;i++){ 58 scanf("%d%d",&u,&v); 59 V[u].push_back(node(v,i)); 60 V[v].push_back(node(u,i)); 61 } 62 for(int i=1;i<=n;i++){ 63 if(!vis[i]){ 64 dis[i]=0; 65 Tarjan(i); 66 mark[i]=1; 67 } 68 } 69 //for(int i=1;i<=n;i++) printf(i==n?"%d\n":"%d ",dis[i]); 70 //for(int i=1;i<=n;i++) printf(i==n?"%d\n":"%d ",fa[i]); 71 for(int i=1;i<=c;i++){ 72 if(res[i]==-1) puts("Not connected"); 73 else printf("%d\n",res[i]); 74 } 75 } 76 }
