### nvson2201's blog

By nvson2201, history, 3 months ago,

If i have q query [start, end] in the graph. How to check whether:

the length of start to end can be arbitrarily negative length?

0  floydWarshall(matrix, n) // RUN THE FLOYD WARSHALL ALGORITHM
1  FOR (start, end) in q:
2  check = false
2  while(start != end)
3     start = Next(start)
4     if (dist[start][start] < 0)
5          check = true, break
6  if check == true
7     return "Negative Loop here"

I try the code like above and get Time limit because I have a while loop inside which is to find negative loop in the path from start to end.

Can you help me find another method?

• -11

By nvson2201, history, 3 months ago,

# Problem Expedition

A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit of distance it travels.

To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop).

The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000).

Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all.

Input The first line of the input contains an integer t representing the number of test cases. Then t test cases follow. Each test case has the follwing form:

• Line 1: A single integer, N
• Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop.
• Line N+2: Two space-separated integers, L and P

Output For each test case, output a single integer giving the minimum number of fuel stops necessary to reach the town. If it is not possible to reach the town, output -1.

Example Input: 1

4

4 4

5 2

11 5

15 10

25 10

Output:

2

Input details The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively.

Output details: Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.

# My attempt

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

int run_case() {
int N;
cin >> N;
vector<pair<int, int>> biasDistance;
biasDistance.emplace_back(0, 0); // location of tower
priority_queue<pair<int, int>, vector<pair<int, int>>, less<>> biasFuel;
for (int i = 0; i < N; ++i) {
int a, b;
cin >> a >> b;
biasDistance.emplace_back(a, b);
}
sort(biasDistance.begin(), biasDistance.end());

int L, P;
cin >> L >> P;

int res = 0;
while (!biasDistance.empty()) {
// which points we can reach
while (P >= L - biasDistance.back().first && !biasDistance.empty()) {
biasFuel.push({biasDistance.back().second, biasDistance.back().first});
biasDistance.pop_back();
}

if (biasDistance.empty()) { // if we can reach all the points
return res;
} else { // if not enough fuel to next point --> need stop somewhere
while (P < L - biasDistance.back().first && !biasFuel.empty()) {
int addedFuel = biasFuel.top().first;
int location = biasFuel.top().second;
biasFuel.pop();
if (L >= location) {
P = P - (L - location) + addedFuel;
L = location;
} else {
}
res++; // stop here
}
}

// still not enough fuel to next point
if (P < L - biasDistance.back().first) {
return -1;
}
}

return res;
}

int main() {
int T;
cin >> T;
while (T > 0) {
cout << run_case() << "\n";
T--;
}
}


# Request for help

I tried to spend a lot of time to solve this problem. But i failed almost tests with the answer is -1

Don't know why. Could you help me?

• 0

By nvson2201, history, 4 months ago,

https://codeforces.com/contest/1418/submission/92798564

#include <algorithm>
#include <array>
#include <cassert>
#include <iostream>
#include <vector>
using namespace std;

template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << '{'; string sep; for (const auto &x : v) os << sep << x, sep = ", "; return os << '}'; }
template<typename T, size_t size> ostream& operator<<(ostream &os, const array<T, size> &arr) { os << '{'; string sep; for (const auto &x : arr) os << sep << x, sep = ", "; return os << '}'; }
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }

#ifdef NEAL_DEBUG
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

const int INF = 1e9 + 5;

void run_case() {
int N;
cin >> N;
vector<int> A(N);

for (auto &a : A)
cin >> a;

vector<array<int, 2>> dp(N + 1, {INF, INF});
dp[0][1] = 0;

for (int i = 0; i < N; i++)
for (int who = 0; who < 2; who++)
for (int fight = 1; fight <= min(N - i, 2); fight++) {
int hard = A[i] + (fight > 1 ? A[i + 1] : 0);
dp[i + fight][!who] = min(dp[i + fight][!who], dp[i][who] + who * hard);
}

cout << min(dp[N][0], dp[N][1]) << '\n';
}

int main() {
ios::sync_with_stdio(false);
#ifndef NEAL_DEBUG
cin.tie(nullptr);
#endif

int tests;
cin >> tests;

while (tests-- > 0)
run_case();
}


can you help me explain these code?