제출 #777270

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
777270		bachhoangxuan	Soccer (JOI17_soccer)	C++17	35 / 100	317 ms	38696 KiB

soccer

// Judges with GCC >= 12 only needs Ofast
// #pragma GCC optimize("O3,no-stack-protector,fast-math,unroll-loops,tree-vectorize")
// MLE optimization
// #pragma GCC optimize("conserve-stack")
// Old judges
// #pragma GCC target("sse4.2,popcnt,lzcnt,abm,mmx,fma,bmi,bmi2")
// New judges. Test with assert(__builtin_cpu_supports("avx2"));
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma,tune=native")
// Atcoder
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma")
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
- insert(x),erase(x)
- find_by_order(k): return iterator to the k-th smallest element
- order_of_key(x): the number of elements that are strictly smaller
*/
#include<bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_real_distribution<> pp(0.0,1.0);
#define int long long
#define ld long double
#define pii pair<int,int>
#define piii pair<pii,int>
#define mpp make_pair
#define fi first
#define se second
const int inf=1e18;
const int mod=1e9+7;
const int maxn=1005;
const int bl=650;
const int maxs=655;
const int maxm=200005;
const int maxq=500005;
const int maxl=20;
const int maxa=250000;
const int root=3;
int power(int a,int n){
    int res=1;
    while(n){
        if(n&1) res=res*a%mod;
        a=a*a%mod;n>>=1;
    }
    return res;
}
const int iroot=power(3,mod-2);
const int base=101;
struct dd{
    int x,y,t,d;
    friend bool operator<(dd a,dd b){return a.d<b.d;}
    friend bool operator>(dd a,dd b){return a.d>b.d;}
};
int dx[]={0,0,1,-1},
    dy[]={1,-1,0,0};
int n,m,a,b,c,sz,f[maxn][maxn],dist[maxn][maxn][10];
void solve(){
    cin >> n >> m >> a >> b >> c >> sz;
    vector<pii> p(sz+1);
    memset(f,-1,sizeof(f));
    queue<pii> q;
    for(int i=1;i<=sz;i++){
        cin >> p[i].fi >> p[i].se;
        f[p[i].fi][p[i].se]=0;q.push(p[i]);
    }
    auto ok = [&](int x,int y){
        if(x<0 || y<0 || x>n || y>m) return false;
        return true;
    };
    while(!q.empty()){
        auto &[x,y]=q.front();q.pop();
        for(int t=0;t<4;t++){
            int xt=x+dx[t],yt=y+dy[t];
            if(ok(xt,yt) && f[xt][yt]==-1) f[xt][yt]=f[x][y]+1,q.push({xt,yt});
        }
    }
    /*
    for(int i=0;i<=n;i++){
        for(int j=0;j<=m;j++) cout << f[i][j] << ' ';
        cout << '\n';
    }
    */
    for(int i=0;i<=n;i++) for(int j=0;j<=m;j++) for(int t=0;t<=4;t++) dist[i][j][t]=inf;
    priority_queue<dd,vector<dd>,greater<dd>> pq;
    dist[p[1].fi][p[1].se][4]=0;pq.push({p[1].fi,p[1].se,4,0});;
    while(!pq.empty()){
        auto [x,y,dir,d]=pq.top();pq.pop();
        if(d!=dist[x][y][dir]) continue;
        //cout << x << ' ' << y << ' ' << dir << ' ' << d << '\n';
        if(dir==4){
            for(int t=0;t<4;t++){
                int xt=x+dx[t],yt=y+dy[t];
                if(!ok(xt,yt)) continue;
                if(dist[xt][yt][t]>d+a+b){
                    dist[xt][yt][t]=d+a+b;
                    pq.push({xt,yt,t,d+a+b});
                }
                if(dist[xt][yt][4]>d+c){
                    dist[xt][yt][4]=d+c;
                    //cout << xt << ' ' << yt << ' ' << 4 << ' ' << d+c << '\n';
                    pq.push({xt,yt,4,d+c});
                }
            }
        }
        else{
            int xt=x+dx[dir],yt=y+dy[dir];
            if(ok(xt,yt) && dist[xt][yt][dir]>d+a){
                dist[xt][yt][dir]=d+a;
                pq.push({xt,yt,dir,d+a});
            }
            if(dist[x][y][4]>d+f[x][y]*c){
                dist[x][y][4]=d+f[x][y]*c;
                pq.push({x,y,4,d+f[x][y]*c});
            }
        }
    }
    cout << dist[p[sz].fi][p[sz].se][4] << '\n';
}
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    int test=1;//cin >> test;
    while(test--) solve();
}

• Subtask 1: 5 / 5
• Subtask 2: 30 / 30
• Subtask 3: 0 / 65