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#284085#5458. Shortest Path Querybobbilyking0 2ms5792kbC++173.0kb2023-12-16 05:00:552023-12-16 05:00:57

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你现在查看的是测评时间为 2023-12-16 05:00:57 的历史记录

  • [2024-06-21 12:38:30]
  • 管理员手动重测本题所有提交记录
  • [2023-12-16 05:00:57]
  • 评测
  • 测评结果:0
  • 用时:2ms
  • 内存:5792kb
  • [2023-12-16 05:00:55]
  • 提交

answer

#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

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

typedef int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)

#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)

#define NN 50010
#define M 1000000007 // 998244353

vector<pl> opt[NN];
ll indeg[NN];
vector<pl> adj[NN];

int main(){
//    freopen("a.in", "r", stdin);
//    freopen("a.out", "w", stdout);

    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(20);
    G(n) G(m) while(m--) {
        G(u) G(v) G(c)
        adj[u].emplace_back(v,c);
        indeg[v]++;
    }
    opt[1] = {pair(0, 0)};
    {
        vector<ll> q{1}, nq;
        for (; q.size(); swap(q, nq), nq.clear()) {
            for (auto i: q) {
                for (auto [x, w]: adj[i]) {
                    vector<pl> temp;
                    // assume opt's are convex hulls in sorted order already
                    // then we simply need to do a linear merge
                    ll lp = 0, rp = 0;

                    auto append = [&](ll x, ll y) {
                        while (temp.size() and y <= temp.back().V) temp.pop_back();
                        temp.emplace_back(x, y);
                    };

                    while (lp < opt[i].size() and rp < opt[x].size()) {
                        auto [x1, y1] = opt[i][lp];
                        x1 += (w==0); y1 += (w==1);
                        auto [x2, y2] = opt[x][rp];
                        if (x1 >= x2 and y1 >= y2) { // merge left first
                            append(x1, y1), lp++;
                        } else {
                            append(x2, y2), rp++;
                        }
                    }
                    while (lp < opt[i].size()) {
                        auto [x1, y1] = opt[i][lp];
                        x1 += (w==0); y1 += (w==1);
                        append(x1, y1); lp++;
                    }
                    while (rp < opt[x].size()) {
                        auto [x2, y2] = opt[x][rp];
                        append(x2, y2), rp++;
                    }

                    opt[x] = temp;
                    if (!--indeg[x]) nq.push_back(x);
                }
            }
        }
    }

    F(i, 1, n+1) {
        // The issue with this pareto front scan is that it's not the 
        cout << i << ": " << opt[i].size() << endl;
        F(j, 1, opt[i].size()) assert(opt[i][j-1].K > opt[i][j].K and opt[i][j-1].V < opt[i][j].V);
    }

    exit(0);

    G(q) while(q--){
        G(a) G(b) G(t)
        ll ans = INT_MAX;
        for (auto [x, y]: opt[t]) ans = min(ans, x*a + y*b);
        cout << ans << '\n';
    }
    
}

详细

Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 5792kb

input:

4 4
1 2 0
1 3 1
2 4 0
3 4 1
3
3 5 2
3 2 4
2 3 4

output:

1: 1
2: 1
3: 1
4: 2

result:

wrong output format Expected integer, but "1:" found