HDU 2874 LCA在线算法RMQ

Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11256    Accepted Submission(s): 2628

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 21 3 22 4 35 2 31 44 5

 

Sample Output

Not connected6
Hint
HintHuge input, scanf recommended.

解题思路

1，离线的trajan不能做，要查询的太多，都存下来离线处理的话就内存炸了

2，然后就是LCA转RMQ的模板题了，因为要判断不联通，并查集搞一下。

#include<bits/stdc++.h>using namespace std;const int maxn = 100005 ;struct Edge {    int form,to,dist;};vector<Edge> edges ;vector<int>G[maxn] ;bool vis[maxn] ;///标记数组int len[maxn] ;///节点到根的长度int ver[maxn<<1] ;///遍历的节点序列int first[maxn] ;///第一次出现的位置int R[maxn<<1] ;///遍历的深度int d[maxn][50] ;///dp数组，里面放的是RMQ的位置，不是数值int f[maxn];int n,tot ;int find(int x){    return x==f[x]?x:f[x]=find(f[x]) ;}void add_edge(int from,int to,int dist) {    edges.push_back((Edge) {from,to,dist}) ;    int mm = edges.size() ;    G[from].push_back(mm-1) ;}void dfs(int u,int dept) {    f[u] = u ;    vis[u] = true ;    ver[tot] = u ;    R[tot] = dept ;    first[u] = tot ;    tot++ ;    for(int i=0; i<G[u].size(); i++) {        if(!vis[edges[G[u][i]].to]) {            len[edges[G[u][i]].to] = len[u]+ edges[G[u][i]].dist ;            dfs(edges[G[u][i]].to,dept+1) ;            f[edges[G[u][i]].to] = find(u) ;            ver[tot] = u ;            R[tot] = dept ;            tot++ ;        }    }}void RMQ_init(){    for(int i=0;i<tot;i++){///初始化，存的是位置        d[i][0] = i ;    }    for(int j=1;(1<<j)<=tot;j++){        for(int i=0;i+(1<<j)-1<tot;i++){///取最深度最小时的位置            int a = d[i][j-1] ;            int b = d[i+(1<<(j-1))][j-1] ;            if(R[a]<R[b])d[i][j] = d[i][j-1] ;            else d[i][j] = d[i+(1<<(j-1))][j-1] ;        }    }}int RMQ(int l,int r){///返回范围最小深度的位置    if(l>r)return 0;    int k=0;    while((1<<(k+1)) <= r-l+1) k++;    int a = d[l][k] ;    int b = d[r-(1<<k)+1][k] ;    if(R[a]<R[b])return a ;    else return b ;}int LCA(int u ,int v)  ///返回点u和点v的LCA{    int x = first[u] , y = first[v];    if(x > y) swap(x,y);    int res = RMQ(x,y);    return ver[res];}int main() {    int m,c;    while(~scanf("%d%d%d",&n,&m,&c)) {        memset(vis,false,sizeof(vis)) ;        memset(d,false,sizeof(d)) ;        memset(G,false,sizeof(G)) ;        edges.clear() ;        for(int i=0; i<m; i++) {            int u,v,w;            scanf("%d%d%d",&u,&v,&w) ;            u--;v--;            add_edge(u,v,w) ;            add_edge(v,u,w) ;        }        tot = 0;        for(int i=0;i<n;i++){            if(!vis[i]){                dfs(i,1) ;            }        }        RMQ_init() ;        for(int i=0; i<c; i++) {            int a,b;            scanf("%d%d",&a,&b) ;            a--;b--;            if(find(a)!=find(b)){                printf("Not connected\n");                continue ;            }            int ans=len[a]+len[b]-2*len[LCA(a,b)];            printf("%d\n",ans) ;        }    }    return 0;}

Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11256    Accepted Submission(s): 2628

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 21 3 22 4 35 2 31 44 5

 

Sample Output

Not connected6
Hint
HintHuge input, scanf recommended.