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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#288489#5669. Traveling in Jade CityVinhLuuRE 0ms34264kbC++237.5kb2023-12-22 19:17:442023-12-22 19:17:45

Judging History

你现在查看的是最新测评结果

  • [2023-12-22 19:17:45]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:34264kb
  • [2023-12-22 19:17:44]
  • 提交

answer

#include <bits/stdc++.h>
//#define int long long
//#define pb push_back
using namespace std;

typedef pair<int,int> pii;
typedef tuple<int,int,int> tp;
const int N = 6e5 + 5;
const int oo = 2e8;

char Q[N];
int U[N], V[N];

int n, mx, k, m, q, gc[N], gx[N], f[N], c[N], x[N], sum, s[N], R[N], g[N], TX, t[N], NG;
vector<tp> p[N];

int st[5][N << 2], lz[5][N << 2];

void dolz(int j,int i){
    if(lz[j][i] == 0) return;
    int x = lz[j][i];
    lz[j][i] = 0;
    lz[j][i << 1] += x;
    lz[j][i << 1|1] += x;
    st[j][i << 1] += x;
    st[j][i << 1|1] += x;
}

void update(int j,int i,int l,int r,int u,int v,int x){
    if(l > r || l > v || r < u) return;
    if(u <= l && r <= v) {
        st[j][i] += x;
        lz[j][i] += x;
        return;
    }
    dolz(j, i);
    int mid = (l + r) >> 1;
    update(j, i << 1, l, mid, u, v, x);
    update(j, i << 1|1, mid + 1, r, u, v, x);
}

int get(int j,int i,int l,int r,int u){
    if(l > r || l > u || r < u || u == 0 || u == n) return 0;
    if(l == r) return st[j][i];
    dolz(j, i);
    int mid = (l + r) >> 1;
    return get(j, i << 1, l, mid, u) + get(j, i << 1|1, mid + 1, r, u);
}

int C(int u,int v){
    if(u == v) return 0;
    if(u > v) swap(u, v);
    int L1 = (get(0, 1, 1, mx, v - 1) - get(0, 1, 1, mx, u - 1) != 0 ? oo : s[v] - s[u]);
    int L2 = (g[n] || NG - (get(0, 1, 1, mx, v - 1) - get(0, 1, 1, mx, u - 1)) != 0 ? oo : sum - (s[v] - s[u]));
//    cout << u << " " << v << " " << L1 << " " << L2 << " " << get(0, 1, 1, mx, u - 1) << " r\n";
    return min(L1, L2);
}

int X(int u,int v){
    if(u == v) return 0;
    if(R[u] > R[v]) swap(u, v);
    if(u == 1 && v == k){
        if(gx[0] || gx[m]) return oo;
        if(get(1, 1, 1, mx, n + m - 1) - get(1, 1, 1, mx, n) != 0) return oo;
        return TX;
    }
    if(u == 1){
        if(get(1, 1, 1, mx, v - 1) - get(1, 1, 1, mx, n) != 0 || gx[0] == true) return oo;
        return t[v] - t[u];
    }
    if(v == k){
        if(get(1, 1, 1, mx, n + m - 1) - get(1, 1, 1, mx, u - 1) != 0 || gx[m] == true) return oo;
        return t[v] - t[u];
    }
    if(get(1, 1, 1, mx, v - 1) - get(1, 1, 1, mx, u - 1) != 0) return oo;
    return t[v] - t[u];
}

void sub3(){

    mx = n + m;

    for(int i = 2; i <= n; i ++) s[i] = s[i - 1] + c[i - 1];

    R[1] = 0;
    int cnt = 0;
    for(int i = n + 1; i <= n + m; i ++){
        g[i] = 1;
        R[i] = ++cnt;
        t[i] = t[i - 1] + x[i - n - 1];
    }
    R[k] = ++cnt;
    t[k] = t[n + m] + x[m];

    for(int j = 1; j <= q; j ++){
        char type = Q[j];
        if(type == 'q'){
            int u = U[j], v = V[j];
            int kq = oo;
            if(g[u] == g[v]){
                if(g[u] == 1){
                    kq = min({X(u, v), X(u, 1) + X(v, k) + C(1, k), C(1, k) + X(u, k) + X(v, 1)});
                }else{
                    kq = min({C(u, v), C(1, u) + C(k, v) + X(1, k), C(k, u) + C(1, v) + X(1, k)});
                }
            }else{
                if(g[u] == 1){
                    kq = min({X(1, u) + C(1, v), X(k, u) + C(k, v)}) ;
                }else{
                    kq = min({X(k, v) + C(k, u), X(1, v) + C(1, u)});
                }
            }
            if(kq >= oo) cout << "impossible\n";
            else cout << kq << "\n";
        }else if(type == 'c'){
            int i = U[j];
            if(gc[i] == 0){
                NG++;
                update(0, 1, 1, mx, i, n, 1);
                gc[i] = 1;
            }else{
                NG--;
                gc[i] = 0;
                update(0, 1, 1, mx, i, n, -1);
            }
        }else{
            int i = U[j];
            if(gx[i] == 0){
                if(i != 0 && i != m) update(1, 1, 1, mx, n + i, n + m, 1);
                gx[i] = 1;
            }else{
                gx[i] = 0;
                if(i != 0 && i != m) update(1, 1, 1, mx, n + i, n + m, -1);
            }
        }
    }
}

int dij(int u,int v){
    for(int i = 1; i <= n + m; i ++) f[i] = oo;
    priority_queue<pii,vector<pii>,greater<pii>> pq;
    pq.push({f[u] = 0, u});
    while(!pq.empty()){
        int w, u; tie(w, u) = pq.top();
        pq.pop();
        if(w > f[u]) continue;
        for(auto tt : p[u]){
            int j, v, e; tie(j, v, e) = tt;
            if(e == 0){
                if(gc[v] == 1) continue;
                if(f[j] > w + c[v]) pq.push({f[j] = w + c[v], j});
            }else{
                if(gx[v] == 1) continue;
                if(f[j] > w + x[v]) pq.push({f[j] = w + x[v], j});
            }
        }
    }
    if(f[v] == oo) return -1;
    else return f[v];
}

void sub1(){
    while(q--){
        char type; cin >> type;
        if(type == 'q'){
            int u, v; cin >> u >> v;
            int what = dij(u, v);
            if(what == -1) cout << "impossible\n";
            else cout << what << "\n";
        }else if(type == 'c'){
            int i; cin >> i;
            gc[i] = 1 - gc[i];
        }else{
            int i; cin >> i;
            gx[i] = 1 - gx[i];
        }
    }
}

int calc(int u,int v){
    if(u == v) return 0;
    if(u > v) swap(u, v);
    return min(s[v] - s[u], sum - (s[v] - s[u]));
}

int calx(int u,int v){
    if(u == v) return 0;
    if(R[u] > R[v]) swap(u, v);
    return t[v] - t[u];
}

void sub2(){

    for(int i = 2; i <= n; i ++) s[i] = s[i - 1] + c[i - 1];

    R[1] = 0;
    int cnt = 0;
    for(int i = n + 1; i <= n + m; i ++){
        g[i] = 1;
        R[i] = ++cnt;
        t[i] = t[i - 1] + x[i - n - 1];
    }
    R[k] = ++cnt;
    t[k] = t[n + m] + x[m];

   for(int i = 1; i <= q; i ++){

        char type = Q[i];
        int u = U[i], v = V[i];
        if(g[u] == g[v]){
            if(g[u] == 1){
                cout << min({calx(u, v), calx(u, 1) + calx(v, k) + calc(1, k), calc(1, k) + calx(u, k) + calx(v, 1)}) << "\n";
            }else{
                cout << min({calc(u, v), calc(1, u) + calc(k, v) + TX, calc(k, u) + calc(1, v) + TX}) << "\n";
            }
        }else{
            if(g[u] == 1){
                cout << min({calx(1, u) + calc(1, v), calx(k, u) + calc(k, v)}) << "\n";
            }else{
                cout << min({calx(k, v) + calc(k, u), calx(1, v) + calc(1, u)}) << "\n";
            }
        }
    }
}


signed main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    if(fopen("train.inp","r")){
        freopen("train.inp","r",stdin);
        freopen("train.out","w",stdout);
    }
    cin >> n >> k >> m >> q;
    for(int i = 1; i <= n; i ++){
        cin >> c[i];
        sum += c[i];
        #define pb push_back
        p[i].push_back({i % n + 1, i, 0});
        p[i % n + 1].pb({i, i, 0});
    }
    for(int i = 0; i <= m; i ++){
        cin >> x[i];
        TX += x[i];
        if(i == 0){
            p[1].pb({n + 1, i, 1});
            p[n + 1].pb({1, i, 1});
        }
        else if(i == m){
            p[n + m].pb({k, i, 1});
            p[k].pb({n + m, i, 1});
        }
        else{
            p[n + i].pb({n + i + 1, i, 1});
            p[n + i + 1].pb({n + i, i, 1});
        }
    }
//    if(n <= 2000 && q <= 2000 && k <= 2000 && m <= 2000){
//        sub1();
//        return 0;
//    }
    bool ff = true;
    for(int i = 1; i <= q; i ++){
        cin >> Q[i];
        if(Q[i] == 'q') cin >> U[i] >> V[i];
        else{
            ff = false;
            cin >> U[i];
        }
    }
    if(ff) sub2();
    else sub3();
}


详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 34264kb

input:

4 3 1 9
2 3 8 4
1 1
q 3 4
x 0
q 3 4
c 3
q 3 4
c 1
q 3 4
x 0
q 3 4

output:

6
8
9
impossible
6

result:

ok 5 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 32500kb

input:

4 3 1 9
2 3 8 4
1 1
q 3 4
x 0
q 3 4
c 3
q 3 4
c 1
q 3 4
x 0
q 3 4

output:

6
8
9
impossible
6

result:

ok 5 lines

Test #3:

score: -100
Runtime Error

input:

1000000 999999 1000000 1000000
200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 2...

output:


result: