結果

問題 No.2630 Colorful Vertices and Cheapest Paths
ユーザー woodywoodywoodywoody
提出日時 2024-02-17 01:03:44
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 360 ms / 2,500 ms
コード長 5,362 bytes
コンパイル時間 3,484 ms
コンパイル使用メモリ 233,792 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-09-28 23:19:32
合計ジャッジ時間 9,012 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
(要ログイン)

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 109 ms
6,820 KB
testcase_01 AC 253 ms
6,816 KB
testcase_02 AC 306 ms
6,816 KB
testcase_03 AC 2 ms
6,816 KB
testcase_04 AC 2 ms
6,816 KB
testcase_05 AC 2 ms
6,820 KB
testcase_06 AC 118 ms
6,816 KB
testcase_07 AC 290 ms
6,816 KB
testcase_08 AC 278 ms
6,816 KB
testcase_09 AC 298 ms
6,816 KB
testcase_10 AC 352 ms
6,816 KB
testcase_11 AC 351 ms
6,816 KB
testcase_12 AC 356 ms
6,816 KB
testcase_13 AC 352 ms
6,816 KB
testcase_14 AC 360 ms
6,816 KB
testcase_15 AC 211 ms
6,816 KB
testcase_16 AC 210 ms
6,820 KB
testcase_17 AC 209 ms
6,820 KB
testcase_18 AC 114 ms
6,816 KB
testcase_19 AC 167 ms
6,820 KB
testcase_20 AC 152 ms
6,816 KB
testcase_21 AC 191 ms
6,816 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto& i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll

using namespace std;
using namespace atcoder;

using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}}
#define retval(x) {cout << #x << endl; return;}
#define cout2(x,y) cout << x << " " << y << endl;
#define coutp(p) cout << p.fi << " " << p.se << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define assertmle(x) if (!(x)) {vi v(3e8);}
#define asserttle(x) if (!(x)) {while(1){}}
#define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define coutv2(v) fore(vy , v) cout << vy << "\n";
#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define coutvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = dynamic_modint<0>;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define no {cout <<"No"<<endl;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#include "./debug/localDebug.cpp"
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show1(a) cerr<<#a<<" = "<<a<<endl
#define show2(a,b) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<endl
#define show3(a,b,c) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<endl
#define show4(a,b,c,d) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<endl
#define show5(a,b,c,d,e) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<" : "<<#e<<" = "<<e<<endl
#define DEBUG_LINE cout << "DEBUG_LINE : " << __LINE__ << endl
#define showv(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) {cout<<#v<<endl; fore(vy , v) cout << vy << "\n";}
#define showvm(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) {cout<<#v<<endl; fore(vy , v) cout << vy.val() << "\n";}
#define showmat(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg << " "; cout << endl;}}
#define showmatm(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg.val() << " "; cout << endl;}}
#else
#define showp(p)
#define show1(a)
#define show2(a,b)
#define show3(a,b,c)
#define show4(a,b,c,d)
#define show5(a,b,c,d,e)
#define DEBUG_LINE
#define showv(v)
#define showv2(v)
#define showvm(v)
#define showvm2(v)
#define showmat(v)
#define showmatm(v)
#endif
#define overload5(a,b,c,d,e,f, ...) f
#define show(...) overload5(__VA_ARGS__, show5, show4, show3, show2, show1)(__VA_ARGS__)

void solve(){
    int n,m; cin>>n>>m;
    vvi e(n);
    rep(i,m){
        int a,b; cin>>a>>b;
        a--; b--;
        e[a].pb(b);
        e[b].pb(a);
    }
    vi c(n);
    rep(i,n) cin>>c[i],c[i]--;

    vl w(10); rep(i,10) cin>>w[i];
    int l =  1<< 10;

    int q; cin>>q;
    vi a(q),b(q);
    rep(i,q){
        cin>>a[i]>>b[i];
        a[i]--; b[i]--;
    }

    vl sum(l);
    vl ans(q,LINF);
    rep(i,l){
        dsu d(n);
        rep(j,n){
            if ((i>>c[j]&1)==0) continue;
            fore(y, e[j]){
                if ((i>>c[y]&1)==0) continue;
                d.merge(j, y);
            }
        }
        rep(j,10) if (i>>j&1) sum[i] += w[j];
        rep(j,q){
            if (d.same(a[j], b[j])) chmin(ans[j], sum[i]);
        }
    }
    rep(i,q) if (ans[i]==LINF) ans[i] = -1;
   
    outv;


    return;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t = 1;
    //cin>>t;

    rep(i,t){
        solve();
    }

    return 0;
}
0