Submission #756130

#	Time	Username	Problem	Language	Result	Execution time	Memory
756130		HaciyevAlik	Cyberland (APIO23_cyberland)	C++17	44 / 100	936 ms	8800 KiB

cyberland

#include "cyberland.h"
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define ff first
#define ss second
const int mx=1e5+5;
int n,m,h,k,used[mx],used2[15];
long dis[mx];
vector<pair<int,int>> g[mx];
struct triple {
    int u,v,w;
};
bool comp(triple a,triple b) {
    if(a.u < b.u) {
        return 0;
    }
    return 1;
}
void dfs(int u) {
    used[u]=1;
    for(int i=0;i<int(g[u].size());++i) {
        int v=g[u][i].ff;
        if(!used[v]&&v!=h) dfs(v);
    }
}
void dfs2(int u) {
    used2[u]=1;
    for(int i=0;i<int(g[u].size());++i) {
        int v=g[u][i].ff;
        if(!used2[v]) dfs2(v);
    }
}
double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> arr) {
    n=N,m=M,h=H,k=K;
    for(int i=0;i<mx;++i) g[i].clear();
    memset(used,0,sizeof(used));
    for(int i=0;i<m;++i) {
        g[x[i]].pb({y[i],c[i]});
        g[y[i]].pb({x[i],c[i]});
    }
    //check if it's a chain and apply all the special abilities -> m=n-1 && x[i]=i && y[i]=i+1
    if(m==n-1) {
        bool f=1;
        vector<triple> v;
        for(int i=0;i<m;++i) {
            triple cur;
            cur.u = x[i],cur.v=y[i],cur.w=c[i];
            v.pb(cur);
            if(x[i]+1!=y[i]) {
                f=0; break;
            }
        }
        if(f) {
            sort(v.begin(),v.end(),comp);
            double res=0.0;
            for(int i=0;i<int(v.size());++i) {
                triple cur=v[i];
                res+=cur.w*1.0;
                if(cur.v==h) {
                    break;
                }
                if(arr[cur.v]==2&&K>0) {
                    res/=2;
                    --K;
                } else if(!arr[cur.v]) {
                    res=0.0;
                }
            }
            return res;
        }
    }
    if(n==2||n==3) {
        memset(used2,0,sizeof(used2));
        dfs2(0);
        if(!used2[h]) {
            return -1;
        }
        if(g[0].size()==1) {
            if(g[0][0].ff==h) {
                return g[0][0].ss;
            } else {
                double res=g[0][0].ss;
                int cur=g[0][0].ff;
                if(!arr[cur]) {
                    res=0;
                } else if(arr[cur]==2 && k>0) {
                    res/=2;
                }
                res+=g[h][0].ss;
                return res;
            }
        } else {
            if(g[0][0].ff==h) {
                double res1=g[0][0].ss,res2=g[0][1].ss;
                int cur=g[0][1].ff;
                if(g[cur].size()==1) return res1;
                if(!arr[cur]) {
                    res2=0;
                } else if(arr[cur]==2&&k>0) {
                    res2/=2;
                }
                if(g[h][0].ff==0) {
                    res2+=g[h][1].ss;
                } else {
                    res2+=g[h][0].ss;
                }
                return min(res1,res2);
            } else {
                double res1=g[0][1].ss,res2=g[0][0].ss;
                int cur=g[0][0].ff;
                if(g[cur].size()==1) return res1;
                if(!arr[cur]) {
                    res2=0;
                } else if(arr[cur]==2&&k>0) {
                    res2/=2;
                }
                if(g[h][0].ff==0) {
                    res2+=g[h][1].ss;
                } else {
                    res2+=g[h][0].ss;
                }
                return min(res1,res2);
            }
        }
    }
    set<int> s;
    dfs(0);
    for(int i=0;i<n;++i) {
        if(used[i]&&!arr[i]) s.insert(i);
    }
    memset(used,0,sizeof(used));
    priority_queue<pair<long,int>> pq;
    for(int i=0;i<n;++i) dis[i]=1e10;
    for(auto i:s) {
        pq.push({0,i});
        dis[i]=0;
    }
    pq.push({0,0});
    dis[0]=0;
    while(!pq.empty()) {
        int u=pq.top().ss; pq.pop();
        if(used[u]) continue;
        used[u]=1;
        for(int i=0;i<int(g[u].size());++i) {
            int v=g[u][i].ff,w=g[u][i].ss;
            if(dis[u]+w<dis[v]) {
                dis[v]=dis[u]+w;
                pq.push({-dis[v],v});
            }
        }
    }
    if(dis[h]==1e10) {
        return -1;
    } else {
        return dis[h];
    }
}

• Subtask 1: 0 / 5
• Subtask 2: 8 / 8
• Subtask 3: 13 / 13
• Subtask 4: 0 / 19
• Subtask 5: 7 / 7
• Subtask 6: 16 / 16
• Subtask 7: 0 / 29
• Subtask 8: 0 / 3