제출 #749228

#제출 시각아이디문제언어결과실행 시간메모리
749228b00norp사이버랜드 (APIO23_cyberland)C++17
97 / 100
3058 ms159592 KiB
#include "cyberland.h"
#include <bits/stdc++.h>
using namespace std;

const int NN = 1e5 + 5, KK = 71 ;
vector<array<int, 2> > g[NN];
bool isreachable[NN];
double dp[NN][KK];
int a[NN];

void dfs(int node, int restricted)
{
    isreachable[node] = true;
    for(auto [to, wt]: g[node])
    {
        if(to == restricted)
        {
            isreachable[restricted] = true;
        }
        if(to == restricted || isreachable[to]) continue;
        dfs(to, restricted);
    }
}

double solve(int N, int M, int K, int H, vector<int> x, 
    vector<int> y, vector<int> c, vector<int> arr) 
{
    K = min(K, 70);
    for(int i = 0; i < N; i++)
    {
        a[i] = arr[i];
        isreachable[i] = false;
        g[i].clear();
        for(int j = 0; j <= K; j++)
        {
            dp[i][j] = 1e18;
        }
    }
    for(int i = 0; i < M; i++)
    {
        g[x[i]].push_back({y[i], c[i]});
        g[y[i]].push_back({x[i], c[i]});
    }
    dfs(0, H);
    if(!isreachable[H])
    {
        return -1;
    }
    priority_queue<pair<double, pair<int, int> > > pq;
    dp[0][0] = 0;
    pq.push({-dp[0][0], {0, 0}});
    for(int i = 1; i < N; i++)
    {
        if(isreachable[i] && arr[i] == 0)
        {
            dp[i][0] = 0;
            pq.push({-dp[i][0], {i, 0}});
        }
    }
    while(!pq.empty())
    {
        auto [dis, l] = pq.top();
        auto [node, k] = l;
        // cout << "dis = " << dis << ", node = " << node << ", k = " << k << "\n";
        pq.pop();
        dis = -dis;
        if(dp[node][k] < dis)
        {
            continue;
        }
        for(auto [to, wt]: g[node])
        {
            if(dp[to][k] > dp[node][k] + wt)
            {
                dp[to][k] = dp[node][k] + (long long)wt;
                if(to != H)
                {
                    pq.push({-dp[to][k], {to, k}});
                }
            }
            // cout << "a[to] = " << a[to] << "k = " << k << ", K = " << K << ", dp[to][k + 1] = " << dp[to][k + 1] << "\n";
            if(a[to] == 2 && k != K && dp[to][k + 1] > (dp[node][k] + (double)wt) / 2.0)
            {
                dp[to][k + 1] = (dp[node][k] + wt) / 2.0;
                if(to != H)
                {
                    pq.push({-dp[to][k + 1], {to, k + 1}});
                }
            }
        }
    }
    double ans = 1e18;
    for(int i = 0; i <= K; i++)
    {
        ans = min(ans, dp[H][i]);
    }
    return ans;
}
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