Submission #1113242

#	Time	Username	Problem	Language	Result	Execution time	Memory
1113242		minhvule	Feast (NOI19_feast)	C++17	41 / 100	18 ms	3268 KiB

feast

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define fi first
#define se second
#define pb push_back
#define freo(task) if(fopen(task".inp", "r")){ freopen(task".inp", "r", stdin); freopen(task".out", "w", stdout); }
#define FOR(i, a, b) for(int i = (a), _b = (b); i <= _b; i ++)
#define FORD(i, a, b) for(int i = (a), _b = (b); i >= _b; i --)
#define bit(x, i) ((x >> i) & 1)
#define oo 1e18
#define all(v) v.begin(), v.end()

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
// mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
// ll rand(ll l, ll r) { assert(l <= r); return uniform_int_distribution<ll>(l, r)(rd); }
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int base = 31;
int n, k, a[N], dp[N], s[N];
/*
sub5: dp O(n ^ 3)
dp1[i][j] là tổng lớn nhất khi chia i món cho j thằng
dp1[i][j] = max(dp1[i - 1][j], max(dp1[k - 1][j]) + (a[k] -> i)
*/
int dp1[305][305];
void sub5(){
    FOR(i, 1, n) cin>>a[i], a[i] += a[i - 1];
    FOR(i, 1, n)
        FOR(j, 1, k){
            dp1[i][j] = dp1[i - 1][j];
            FOR(t, 1, i - 1)
                dp1[i][j] = max(dp1[i][j], dp1[t - 1][j - 1] + a[i] - a[t - 1]);
        }
    // FOR(i, 1, n) cout<<dp1[i][k]<<" ";
    cout<<dp1[n][k];
}
int dp2[2005][2005], g[2005][2005];
/*
sub6: O(n ^ 2)
dp[i][j] là tổng lớn nhất khi chia i món cho j thằng vua
g[i][j] là giá tri tối ưu khi chia i món với số lượng vua bất kì từ 1..j
vì có thằng vua có thể không có
*/
void sub6(){
    FOR(i, 1, n) cin>>a[i];
    FOR(i, 1, n)
        FOR(j, 1, k){
            dp2[i][j] = max(dp2[i - 1][j] + a[i], g[i - 1][j - 1] + a[i]);
            g[i][j] = max(g[i - 1][j], dp2[i][j]);
        }
    int ans = 0;
    FOR(j, 1, k) ans = max(ans, g[n][j]);
    cout<<ans;
}
/*
sub7: O(nlog)
tham lam
xoa cac phan tu am o dau va cuoi cua mang
gop cac gia tri am duong -> mang a am duong xen ke
*/
int res[N], pre[N], nxt[N];
bool check[N];
vector<int> v;
priority_queue<tuple<int, int, int>> pq;
void sub7(){
    FOR(i, 1, n) cin>>a[i];
    int l = 1, r = n;
    while(l <= n && a[l] <= 0) l ++;
    while(r >= 1 && a[r] <= 0) r--;
    if(l > r){
        cout<<0;
        return;
    }
    n = 0;
    FOR(i, l, r)
        a[++ n] = a[i];
    // FOR(i, 1, n) cout<<a[i]<<" ";
    FOR(i, 1, n) s[i] = s[i - 1] + a[i];
    int last = 0, cnt = 0;
    FOR(i, 1, n)
        if(i > 1 && (a[i] > 0) != (a[i - 1] > 0)){
            if(a[i - 1] > 0) cnt ++;
            res[i - 1] = -abs(s[i - 1] - s[last]);
            pre[i - 1] = last;
            nxt[last] = i - 1;
            pq.push({res[i - 1], last, i - 1});
            last = i - 1;
        }
    cnt ++;
    res[n] = -abs(s[n] - s[last]);
    pre[n] = last;
    nxt[last] = n;
    pq.push({res[n], last, n});
    int ans = 0;
    while(!pq.empty() && cnt > k){
        int val = get<0>(pq.top()),
        l = get<1>(pq.top()),
        r = get<2>(pq.top());
        pq.pop();
        if(check[l] || check[r] || val != res[r]) continue;
        cnt --;
        check[l] = check[r] = 1;
        ans += res[r];
        int i = pre[l], j = nxt[r];
        if(i < j && i >= 0 && j <= n){
            pre[j] = i, nxt[i] = j;
            res[j] = res[j] + res[l] - res[r];
            res[r] = -oo;
        }
        if(i < j && i >= 0 && j <= n && !check[j]) pq.push({res[j], i, j});
    }
    FOR(i, 1, n) if(a[i] > 0) ans += a[i];
    cout<<ans;
}
void solve(){
    cin>>n>>k;
    sub7();
}
int32_t main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    // freo("");
    int nTest = 1;
    // cin>>nTest;
    while(nTest --) solve();
    // cerr << "\nTime: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
    return 0;
}
/*



*/

• Subtask 1: 0 / 4
• Subtask 2: 0 / 8
• Subtask 3: 0 / 18
• Subtask 4: 10 / 10
• Subtask 5: 11 / 11
• Subtask 6: 20 / 20
• Subtask 7: 0 / 29