590 - Always on the run

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590 - Always on the run

Post by shaikat » Mon Jul 21, 2003 10:29 am

Can the problem-590 be solved by BFS?
If it can then how.
plz explain. :)

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590 can you give me a tiny hint

Post by IBelieve » Sun Nov 16, 2003 7:54 pm

Am I the only one who doesn't know how to do that one ??
When people agree with me I always feel that I must be wrong.

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Post by jackie » Thu Aug 26, 2004 12:22 pm

I just finished it.

You may not consider it as a graph problem.

Just for each day calculate the min cost from paris to other cities.

Surely you can get min cost from paris after first day.

Then for the second day from the results we just calculate it's easy to get the min cost for every city after two days travel.

the third day
the fourth day
the kth day

The min cost to nth city after kth day is the answer.

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Post by Jan » Sun Apr 09, 2006 12:54 am

It can be solved using bfs.

The idea is simple.

Code: Select all

for i = 1 to n // n total places
    for d = 1 to day // day total days
        val[i][d] = inf

set val[1][0] = 0, because in 0-th day she is in the 1st place. And the cost is zero.

1. enque(1, 0)
2. u = deque()
3. for i = 1 to n and i is different from u.n
4.     v.n = i and v.day = u.day+1
5.     if(there is a flight in the u.day-th day from u.n to v.n)
6.         if(val[v.n][v.day] > val[u.n][u.day] + flight cost)
7.             update val[v.n][v.day] and enque(v)
8. if the queue is not empty then goto step 2

then the val[n][day] will contain the minimum cost.
Hope it helps.
Ami ekhono shopno dekhi...

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Location: Dhaka, Bangladesh

Post by Jan » Sun Apr 09, 2006 1:01 am

You can also solve it using bfs.
Ami ekhono shopno dekhi...

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Re: 590 can you give me a tiny hint

Post by bourne » Mon Sep 15, 2008 11:37 pm

I solved this problem using recursion with memoization. But I am getting WA.
Here is my recursive function,

Code: Select all

int n,k;
vector<int> D[11][11]; //D[city1][city2][day]
long long dp[11][31]; //dp[city][day]
long long solve(int city,int day)
		if(city==n) return 0;
		return INF;
	long long& res=dp[city][day];
	if(res!=-1) return res;
	for(int i=1;i<=n;i++)
		if(city==i) continue;
		int sz=D[city][i].size();
		if(D[city][i][day%sz]!=0) res=min(res,D[city][i][day%sz]+solve(i,day+1));
	return res;

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Re: 590 can you give me a tiny hint

Post by Imti » Tue Sep 20, 2011 8:39 pm

If You didn't get it accepted yet ..then it's for u...
long long dp[11][31]; //dp[city][day]
day could be as large as 1000 ..so you should increase size of 2nd dimension of your array..I was also getting WA for this reason..:)

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Re: 590 can you give me a tiny hint

Post by Yousuf » Wed Nov 06, 2013 12:13 am

I am getting WR in this problem. Please help
Here is my code.

Code: Select all

AC :) 

Last edited by Yousuf on Wed Nov 06, 2013 9:59 pm, edited 1 time in total.

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Re: 590 can you give me a tiny hint

Post by brianfry713 » Wed Nov 06, 2013 9:47 pm

Print a blank line after each scenario, including the last one.
Check input and AC output for thousands of problems on uDebug!

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Re: 590 can you give me a tiny hint

Post by Yousuf » Wed Nov 06, 2013 9:56 pm

Thanks brianfry713 I got AC now.

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Re: 590 - Always on the run

Post by jporcelli1120 » Thu Feb 26, 2015 10:45 pm

Any one help with small push in right direction, im getting WA

Code: Select all


Repon kumar Roy
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Re: 590 - Always on the run

Post by Repon kumar Roy » Tue May 05, 2015 1:30 pm

WHY WA ???

Code: Select all

#include <bits/stdc++.h>
using namespace std;

/*------- Constants---- */
#define LMT				12
#define ll				long long
#define ull				unsigned long long
#define mod				1000000007
#define MEMSET_INF		63
#define MEM_VAL			1061109567
#define FOR(i,n)			for( int i=0 ; i < n ; i++ )
#define mp(i,j)			make_pair(i,j)
#define lop(i,a,b)		for( int i = (a) ; i < (b) ; i++)
#define pb(a)			push_back((a))
#define gc				getchar_unlocked
#define PI				acos(-1.0)
#define inf				1<<28
#define lc				((n)<<1)
#define rc				((n)<<1 |1)
#define msg(x)			cout<<x<<endl;
#define EPS				1e-7
/*---- short Cuts ------- */
#define ms(ara_name,value) memset(ara_name,value,sizeof(ara_name))
typedef pair<int, int> ii;
typedef vector<int > vi ;
/*------ template functions ------ */
inline void sc(int &x)
	register int c = gc();
	x = 0;
	int neg = 0;
	for(;((c<48 | c>57) && c != '-');c = gc());
	if(c=='-') {neg=1;c=gc();}
	for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
	if(neg) x=-x;

template <class T> inline T bigmod(T p,T e,T M){
	ll ret = 1;
	for(; e > 0; e >>= 1){
		if(e & 1) ret = (ret * p) % M;
		p = (p * p) % M;
	} return (T)ret;
template <class T> inline T gcd(T a,T b){if(b==0)return a;return gcd(b,a%b);}
template <class T> inline T modinverse(T a,T M){return bigmod(a,M-2,M);}

/*************************** END OF TEMPLATE ****************************/

int N , K ;

int cost[LMT][LMT][1005];
bool flag = false;
int dp[LMT][1005];
int var[LMT];

int cal(int n,int day)
	if( day >= K ){
		if( n == N ) {
			flag = true;
			return 0;
		else return inf ;

	if(dp[n][day] !=-1 ) return dp[n][day];
	int t = inf ;
	for(int i = 1; i <= N ; i ++ ) {
		if(i != n ) {
			if(cost[n][i][day % var[n]] !=0 ) {
				t = min (t , cost[n][i][day % var[n] ] +  cal(i, day + 1) );
	return dp[n][day] = t;

int main()
	int cs = 0 , p ;
	while (1) {
		if(N == 0 && K == 0 ) break;
		ms(cost, 0);
		for(int i = 1 ; i <= N ; i ++ ) {
			for(int j = 1; j <= N ; j ++ ) {
				if( i == j ) continue;
				var[i] = p;
				for(int l  = 0 ; l < p ; l ++ ) {
		ms(dp, -1);
		flag = false;
		int p = cal(1 , 0);
		printf("Scenario #%d\n",++cs);
		if(flag ){
			printf("The best flight costs %d.\n",p);
		else printf("No flight possible.\n");
	return 0;


Sinikar, Uruk, Goran and Kan Falkland islands (malvinas)

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