제출 #1045832

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1045832		becaido	전선 연결 (IOI17_wiring)	C++17	13 / 100	45 ms	14676 KiB

wiring

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,popcnt,sse4,abm")
#include <bits/stdc++.h>
using namespace std;

#ifdef WAIMAI
#include "grader.cpp"
#else
#include "wiring.h"
#endif

#ifdef WAIMAI
#define debug(HEHE...) cout << "[" << #HEHE << "] : ", dout(HEHE)
void dout() {cout << '\n';}
template<typename T, typename...U>
void dout(T t, U...u) {cout << t << (sizeof...(u) ? ", " : ""), dout(u...);}
#else
#define debug(...) 7122
#endif

#define ll long long
#define Waimai ios::sync_with_stdio(false), cin.tie(0)
#define FOR(x,a,b) for (int x = a, I = b; x <= I; x++)
#define pb emplace_back
#define F first
#define S second

const ll INF = 1e18;
const int SIZE = 2e5 + 5;
const int K = 20;

int n, m, sz;
int a[SIZE], b[SIZE];
ll dp[SIZE], pre[SIZE];
pair<int, int> p[SIZE], seg[SIZE];

ll min_total_length(vector<int> _r, vector<int> _b) {
    n = _r.size();
    m = _b.size();
    sz = 0;
    for (int x : _r) p[++sz] = {x, 0};
    for (int x : _b) p[++sz] = {x, 1};
    sort(p + 1, p + sz + 1);
    n = 0;
    for (int i = 1, l = 0; i <= sz; i++) {
        pre[i] = pre[i - 1] + p[i].F;
        if (l == 0) l = i;
        if (i == sz || p[i].S != p[i + 1].S) {
            seg[++n] = {l, i};
            l = 0;
        }
    }
    fill(dp + 1, dp + sz + 1, INF);
    FOR (i, 2, n) {
        auto [l, r] = seg[i];
        auto [pl, pr] = seg[i - 1];
        multiset<ll> ms;
        ll mn = INF;
        FOR (j, pl, pr) ms.insert(dp[j - 1] + 1ll * (pr - j + 1) * p[l].F - (pre[pr] - pre[j - 1]));
        FOR (j, l, r) {
            int cnt = j - l + 1;
            if (pr - pl + 1 >= cnt) mn = min(mn, dp[pr - cnt] + 1ll * cnt * p[pr].F - (pre[pr] - pre[pr - cnt]));
            dp[j] = min(dp[j], mn + (pre[j] - pre[l - 1]) - 1ll * cnt * p[pr].F);
            if (ms.size()) dp[j] = min(dp[j], *ms.begin() + (pre[j] - pre[l - 1]) - 1ll * cnt * p[l].F);
            if (pr - pl + 1 >= cnt) ms.erase(ms.find(dp[pr - cnt] + 1ll * cnt * p[l].F - (pre[pr] - pre[pr - cnt])));
        }
    }
    return dp[sz];
}

/*
in1
4 5
1 2 3 7
0 4 5 9 10
out1
10
*/

• Subtask 1: 0 / 7
• Subtask 2: 13 / 13
• Subtask 3: 0 / 10
• Subtask 4: 0 / 25
• Subtask 5: 0 / 45