priority_queue returns maximum value at each point. But in Dijkstra we want to pick up the smallest possible distance every time. So, when we push the negative distance value into the priority_queue, the maximum value becomes the smallest one. Therefore, each time we can get minimum distance in top of the priority_queue.
Can someone please give me some hints to solve this problem. Thanks in Advance!!
Actually it's a formal binary search approach. If ok function returns false for any mid value that means for taking "mid" maximum gap we can't put C cows. As we know number of cows we can put is inversely proportional to the gap, we have to reduce the maximum gap to put C cows. So we can eliminate mid to upper bound range of the binary search.
Can someone please give me some hints to solve this problem. Thanks in Advance!!
I've updated my code and I think this updated code will give you more intuition. I've replaced the return statement of "ok" function
return (cows < given_cows)
with
return (cows >= given_cows)
Now if "ok" function returns true value then, for the current mid value we can put more than or equal number of cows than given number of cows. As current mid can be the answer, we store mid as answer and increase the lower bound to mid+1. Why? Because we're trying to find the largest X for which (cows >= given_cows) condition is true. If "ok" returns false value, upper bound needs to be updated to mid-1. But why adding 1 with answer? I'm sorry to say that I can't explain it. I did it because of sample example.
For input:
1
This problem would have been a classical problem if the memory limits would not have been that tight. In the classical version we can use DP or the LCS with O(n^2) memory but this problem requires O(n) memory.
How to go about it?
Hints:
Solution:
#include <bits/stdc++.h>using namespace std;#define mxn 5010char str[mxn] , rev[mxn];
Problem in short: Consider an undirected graph of N nodes, M edges. Each edge is labeled with two integers A, B. Find a spanning tree of this graph which has the maximum possible value for the following expression:
#include <bits/stdc++.h>using namespace std;#define mxn 100010struct edge{int u, v, a, b;};edge edg[mxn];int rep[mxn];int n, m, mstA = 0, mstB = 0;void setRep(){for(int i=0; i<mxn; i++) rep[i] = i;}int findRep(int n){if(rep[n]!=n) rep[n] = findRep(rep[n]);return rep[n];}bool connected(int x, int y){
Hints:
Solution:
#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define ff first
Given undirected graph of N nodes, M edges and a ranked list of exciting cities. You are to answer Q queries of the form (u, v, K), asking you to find the cost of the shortest path from u to v...
#include <bits/stdc++.h>using namespace std;#define sfi(x) scanf("%d", &x)#define pb push_back#define mp make_pair#define INF 1000000000#define mxn 155int cost[mxn][mxn];int ranking[mxn];int qsource[6010] , qdest[6010] , ans[6010];vector< int > query[mxn];int main(){int tc, n, c, r, s, d, p, q;sfi(tc);for(int t=1; t<=tc; t++){sfi(n);for(int i=1; i<=n; i++){cost[i][i] = 0;for(int j=i+1; j<=n; j++){sfi(c);
Take as input a list of positive integers. Whenever -1 is given as input, you must output the median of the list, and remove it from the list.
Hints:
Solution:
#include <bits/stdc++.h>using namespace std;#define sf(x) scanf("%d", &x)#define pf(x) printf("%d\n", x)int main(){int n,t;sf(t);
Indian National Olympiad in Informatics (INOI) is round 2 out of 3 (i.e. intermediate) for selection into Indian IOI team.
Indian National Olympiad in Informatics (INOI) is round 2 out of 3 (i.e. intermediate) for selection into Indian IOI team.
