D. Learning to Paint

合集 - 数据结构(28)

1.P3957 [NOIP2017 普及组] 跳房子2024-03-062.P2034 选择数字2024-03-093.P1714 切蛋糕2024-03-094.E - Insert or Erase2024-03-105.P4147 玉蟾宫2024-03-106.P2866 [USACO06NOV] Bad Hair Day S2024-03-107.P2032 扫描2024-03-108.P1884 [USACO12FEB] Overplanting S2024-03-109.P5905 【模板】全源最短路（Johnson）2024-03-1110.P1801 黑匣子2024-03-2911.P2168 [NOI2015] 荷马史诗2024-03-3012.P8306 【模板】字典树2024-03-3013.P1481 魔族密码2024-03-3014.P4341 [BJWC2010] 外星联络2024-04-0115.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0116.P3384 【模板】重链剖分/树链剖分2024-04-0317.P10288 [GESP样题 八级] 区间2024-04-1618.P2161 [SHOI2009] 会场预约2024-04-1719.P1878 舞蹈课2024-05-1720.P9691 [GDCPC2023] Base Station Construction2024-05-1821.P3611 [USACO17JAN] Cow Dance Show S2024-05-2022.F. Cutting Game2024-05-2123.G. Money Buys Less Happiness Now2024-05-2224.D. In Love2024-06-2925.F. Selling a Menagerie2024-06-29

26.D. Learning to Paint2024-07-19

27.P4588 [TJOI2018] 数学计算2024-07-2028.P3522 [POI2011] TEM-Temperature2024-07-21

收起

原题链接

题解

dp+多次优先队列

设 $dp[i]$ 为 $[1,i]$ 区间内，前 $k$ 个最大值（有可能不足k个）(注意 $dp[i]$ 是一个序列)

则 $dp[i]=\{dp[j][t]+a[j+2][i],j\in [0,i-2],t\in [0,to{p}_j]\},\sum t=k$

code

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

int a[1005][1005];

struct Compare {
    bool operator()(const array<int, 4>& a, const array<int, 4>& b) {
        return a[3] < b[3];
    }
};

void solve()
{
    int n, k;
    cin >> n >> k;

    for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j++) cin >> a[i][j];


    vector<vector<int>> dp(n + 5);
    dp[0].push_back(0);

    for (int i = 1; i <= n; i++)
    {
        priority_queue<array<int, 4>, vector<array<int, 4>>, Compare> q;

        q.push({i+1, i - 1, 0, dp[i - 1][0]});

        for (int j = i; j >= 1; j--)  q.push({j, max(0,j - 2), 0, a[j][i] + dp[max(0,j-2)][0]});


        while (!q.empty() && dp[i].size() < k)
        {
            auto [l, it, cnt, val] = q.top();
            q.pop();

            dp[i].push_back(val);
            if (cnt +1< dp[it].size()) q.push({l, it, cnt + 1, dp[it][cnt+1] + a[l][i]});
        }
    }


    for(auto it:dp[n]) cout<<it<<' ';cout<<'\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
}