Tuesday, March 14, 2017

Solution of Light OJ 1257 - Farthest Nodes in a Tree (II)

We may solve using BFS of DFS. Both solutions are given.
Using BFS:
/*  Bismillahir Rahmanir Rahim
    Solution-Using BFS
Find the farthest two nodes and calculate all nodes distance
from both farthest nodes and take the maximum for each node
*/
#include<bits/stdc++.h>
#define fi(n, m) for(int i=n; i<=m; i++)
#define fii(i, n, m) for(int i=n; i<=m; i++)
using namespace std;
vector<int>vt[30001], cost[30001];
int n, dis[30001], dis1[30001], a1, a2, p, cnt=0;

void bfs(int st){
    fi(0, n) dis[i]=-1;
    int mx=-1;
    queue<int>q;
    q.push(st);
    dis[st]=0;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        if(mx<dis[u]){
            mx=dis[u];
            p=u;
        }
        for(int j=0; j<vt[u].size(); j++){
            int v=vt[u][j];
                if(dis[v]==-1){
                    dis[v]=dis[u]+cost[u][j];
                    q.push(v);
                }

        }
    }
}

void bfs1(int st){
    fi(0, n) dis1[i]=-1;
    queue<int>q;
    q.push(st);
    dis1[st]=0;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int j=0; j<vt[u].size(); j++){
            int v=vt[u][j];
                if(dis1[v]==-1){
                    dis1[v]=dis1[u]+cost[u][j];
                    q.push(v);
                }
        }
    }
}

int main()
{
    int t, cs=1, u, v, w;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        fi(1, n-1){
            scanf("%d%d%d", &u, &v, &w);
            vt[u].push_back(v);
            vt[v].push_back(u);
            cost[u].push_back(w);
            cost[v].push_back(w);
        }
        bfs(0); // find an ending point of farthest path
        a1=p;
// find another ending point a2 and distance from a1 (dis[])
        bfs(a1); 
        a2=p;
        bfs1(a2); //calculating distance from a2(dis1[])
        printf("Case %d:\n", cs++);
        fi(0, n-1){
            printf("%d\n", max(dis[i], dis1[i]));
            vt[i].clear(); cost[i].clear();
        }
    }
    return 0;
}


Using DFS:

/*  Bismillahir Rahmanir Rahim
    Solution-Using DFS
Find the farthest two nodes and calculate all nodes distance
from both farthest nodes and take the maximum for each node
*/
#include<bits/stdc++.h>
#define fi(n, m) for(int i=n; i<=m; i++)
#define fii(i, n, m) for(int i=n; i<=m; i++)
using namespace std;
vector<int>vt[30009], cost[30009];
int dis[30009], dis1[30009], vis[30009], n, p, mx;

void dfs(int at){
    if(vis[at]) return ;
    else{
        vis[at]=1;
        if(mx<dis[at]){
            mx=dis[at];
            p=at;
        }
        for(int j=0; j<vt[at].size(); j++){
            int v=vt[at][j];
            if(dis[v]==-1){
                dis[v]=dis[at]+cost[at][j];
                dfs(v);
            }
        }
    }
}

void dfs1(int at){
    if(vis[at]) return ;
    else{
        vis[at]=1;
        for(int j=0; j<vt[at].size(); j++){
            int v=vt[at][j];
            if(dis1[v]==-1){
                dis1[v]=dis1[at]+cost[at][j];
                dfs1(v);
            }
        }
    }
}

int main(){
    int t, cs=1, u, v, w, a1, a2;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        fi(1, n-1){
            scanf("%d%d%d", &u, &v, &w);
            vt[u].push_back(v);
            vt[v].push_back(u);
            cost[u].push_back(w);
            cost[v].push_back(w);
        }
        fi(0, n+1){vis[i]=0; dis[i]=-1;}
        mx=0, dis[0]=0;
        dfs(0); // find an ending point of farthest path
        a1=p, mx=0;
        fi(0, n+1){vis[i]=0; dis[i]=-1;}
        dis[a1]=0;
// find another ending point a2 and distance from a1(dis[])
        dfs(a1);
        a2=p;
        fi(0, n+1){vis[i]=0; dis1[i]=-1;}
        dis1[a2]=0;
        dfs1(a2);//calculating distance from a2(dis1[])
        printf("Case %d:\n", cs++);
        fi(0, n-1){
            printf("%d\n", max(dis[i], dis1[i]));
            vt[i].clear(); cost[i].clear();
        }
    }
    return 0;
}

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