I am trying to solve this problem using BFS. I am not sure what I am doing wrong. Here is the code that I have written: lhUVGD - Online C++0x Compiler & Debugging Tool - Ideone.com
It would be great if someone can help me out.
Thanks!
I am to getting WA in this Question, Please help!!!
problem = AKBAR - Akbar , The great
My Solution → My Submission
please Help!!!
@bishen28 Your code fails for the following simple graph.
1
4 4 1
1 2
2 3
3 4
4 1
1 4
I checked your code, the way you’ve implemented BFS is wrong, in particular, the line
int node = q.front();q.pop();S--;
is problematic.
Decrmenting S for every node in the queue is wrong
What you can change is that each node that goes into queue is a pair, the first one is the node id and the second is strenght left.
So the code would be like this
void BFS(int s,vector<vector<int>> &adj_list,vector<int> &Protected, int S, vector<bool> & visited){
Protected[s]++;
queue<pair<int, int>> q;
visited[s] = true;
q.push({s, S});
while(!q.empty()) {
int node = q.front().first;
int strength_left = q.front().second;
q.pop();
if (strength_left <= 0) {
continue;
}
for(auto adj_node:adj_list[node]){
if(visited[adj_node]==false){
Protected[adj_node]++;
q.push({adj_node, strength_left-1});
visited[adj_node] = true;
}
}
}
}
After making the above changes, you code is giving TLE. I would suggest that you use multi source BFS for this problem and whenever protected[i] gets > 1, you stop.
1 Like
Check my below AC code
// https://www.spoj.com/problems/AKBAR/
#include <bits/stdc++.h>
// Common file
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using ti = tuple<int, int, int>;
using tl = tuple<ll, ll, ll>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pi>;
using vpil = vector<pil>;
using vpli = vector<pli>;
using vpl = vector<pl>;
using vti = vector<ti>;
using vtl = vector<tl>;
using vvi = vector<vi>;
using vvl = vector<vl>;
using vvpi = vector<vpi>;
using vvpil = vector<vpil>;
using vvpli = vector<vpli>;
using vvpl = vector<vpl>;
using vvti = vector<vti>;
using vvtl = vector<vtl>;
using oset = tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;
using omset = tree<pair<int,int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = INT_MAX;
const ll LINF = LLONG_MAX;
const int MOD = 1000000000 + 7;
#define setbits(n) __builtin_popcount(n)
/* #define len(x) (int)size(x) */
/* void setIO(string name = "") { */
/* ios_base::sync_with_stdio(false); */
/* cin.tie(nullptr); */
/* if (len(name)) { */
/* ignore = freopen((name + ".in").c_str(), "r", stdin); */
/* ignore = freopen((name + ".out").c_str(), "w", stdout); */
/* } */
/* } */
const int N = (int) 1e6;
vi g[N+1];
// visited[i] Stores the root node of the bfs tree of which node i is a part
int visited[N+1];
vpi src;
bool bfs() {
queue<pi> q;
int sz = (int) src.size();
for (int i = 0; i < sz; i++) {
q.push({src[i].first, src[i].second});
visited[src[i].first] = src[i].first;
}
while (!q.empty()) {
int u = q.front().first;
int strength = q.front().second;
assert(visited[u]);
q.pop();
if (strength <= 0) {
continue;
}
for (int v : g[u]) {
if (!visited[v]) {
visited[v] = visited[u];
q.push({v, strength-1});
} else {
if (visited[v] == visited[u]) {
continue;
}
assert(visited[v] != visited[u]);
return false;
}
}
}
return true;
}
void reset(int n) {
for (int i = 1; i <= n; i++) {
g[i].clear();
visited[i] = 0;
}
src.clear();
}
int main() {
/* setIO(); */
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T--) {
int n, r, m;
cin >> n >> r >> m;
for (int i = 0; i < r; i++) {
int u, v;
cin >> u >> v;
g[u].emplace_back(v);
g[v].emplace_back(u);
}
/* for (int i = 1; i <= n; i++) { */
/* cout << i << "\n"; */
/* for (int v : g[i]) { */
/* cout << v << " "; */
/* } */
/* cout << endl; */
/* } */
/* cout << endl; */
for (int i = 0; i < m; i++) {
int k, s;
cin >> k >> s;
src.emplace_back(k, s);
}
bool flag = bfs();
if (!flag) {
cout << "No\n";
} else {
bool ans = true;
for (int i = 1; i <= n; i++) {
if (!visited[i]) {
ans = false;
break;
}
}
if (ans) {
cout << "Yes\n";
} else {
cout << "No\n";
}
}
reset(n);
}
return 0;
}
And here is your own AC code with few changes, you can check the changes by using diff, something like vim -d new_code.cc old_code.cc
#include <bits/stdc++.h>
using namespace std;
bool BFS(int s,vector<vector<int>> &adj_list,vector<int> &Protected, int S, vector<bool> & visited){
Protected[s]++;
if (Protected[s] > 1) {
return false;
}
queue<pair<int, int>> q;
visited[s] = true;
q.push({s, S});
while(!q.empty()) {
int node = q.front().first;
int strength_left = q.front().second;
q.pop();
if (strength_left <= 0) {
continue;
}
for(auto adj_node:adj_list[node]){
if(visited[adj_node]==false){
Protected[adj_node]++;
q.push({adj_node, strength_left-1});
visited[adj_node] = true;
}
}
}
return true;
}
void solve(){
int n,r,m;cin>>n>>r>>m;
vector<vector<int>> adj_list(n+1);
vector<int> Protected(n+1,0);
vector<bool> visited(n+1,false);
for(int i=0;i<r;i++){
int A,B;cin>>A>>B;
adj_list[A].push_back(B);
adj_list[B].push_back(A);
}
vector<int> strength(n+1,-1);
for(int i=0;i<m;i++){
int K,S;cin>>K>>S;
strength[K] = S;
}
bool flag = true;
for(int i=1;i<=n;i++){
if(strength[i] >= 0) {
flag = BFS(i,adj_list,Protected,strength[i],visited);
if (!flag) {
break;
}
}
/* for(int s=0;s<=n;s++) visited[s] = false; */
}
/* for(int i=1;i<=n;i++){ */
/* cout << Protected[i] << "\n"; */
/* } */
if (!flag) {
cout << "No\n";
return;
} else {
for(int i=1;i<=n;i++){
if(visited[i] == false) {
flag = false;
break;
}
}
}
if (flag) {
cout << "Yes\n";
} else {
cout << "No\n";
}
return;
}
int main(){
ios_base::sync_with_stdio(0) ;
cin.tie(0) ; cout.tie(0) ;
cout.precision(20);
int T;cin>>T;
while(T--){
solve();
/* if(T)cout << "\n"; */
}
return 0;
}
1 Like
trhe edge vertex array should be 10^7 longer
in que it written R<=min(10^7,N*(N-1)/2) but u make 10^6 longer and ac why ?