Luogu · P1496

火烧赤壁

This C++ solution uses binary search for Luogu P1496 火烧赤壁. Read the reasoning, inspect the code, or try your own test case below.

P1496Sorting & searchingBinary searchC++36 lines
Solution244of 248
Open official problem ↗ Download C++ file ↓ Search the library → Open full Code Lab ↗ Report an issue ↗

Approach

Binary search

火烧赤壁 matches half-open burning intervals and asks for their total union length.

Sorting & searching

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

Open official problem ↗View exact source file ↗
Implementation

P_1496_火烧赤壁.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const long long INF = 1e17; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int n; cin >> n;
        vector<ll> a;
        vector<pair<ll, ll>> v;
        for (int i = 0; i < n; i++) {
            ll x, y; cin >> x >> y;
            a.push_back(x); a.push_back(y);
            v.emplace_back(x, y);
        }
        sort(a.begin(), a.end());
        a.erase(unique(a.begin(), a.end()), a.end());
        auto get_id = [&] (ll x) {
            return lower_bound(a.begin(), a.end(), x) - a.begin();
        };
        vector<bool> vis(a.size() - 1);
        //or use differene array to optimize
        for (int i = 0; i < v.size(); i++) {
            auto[x, y] = v[i];
            x = get_id(x); y = get_id(y);
            //[x, y)
            for (int j = x; j < y; j++) {
                vis[j] = 1;
            }
        }
        ll ans = 0;
        for (int i = 0; i < vis.size(); i++) {
            if (vis[i]) ans += a[i + 1] - a[i];
        }
        cout << ans << '\n';
        return 0;
    }
        

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗Keep studying →

Test this problem

Run your code here.

Paste your code, run a test case, compare the output, or trace selected values.

Full trace, comparison & stress testing ↗
StatusReady
Output
No run yet.
Diagnostics
No diagnostics yet.

Each run is isolated and has strict limits. Passing one test does not guarantee the judge will accept the solution.