D. World is Mine

合集 - 动态规划dp(55)

1.P8784 [蓝桥杯 2022 省 B] 积木画2024-02-172.P9847 [ICPC2021 Nanjing R] Crystalfly2024-02-183.G. Vlad and Trouble at MIT2024-02-204.P2899 [USACO08JAN] Cell Phone Network G2024-02-205.P5322 [BJOI2019] 排兵布阵2024-03-036.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-037.P1156 垃圾陷阱2024-03-038.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-049.P2466 [SDOI2008] Sue 的小球2024-03-0410.P5020 [NOIP2018 提高组] 货币系统2024-03-0411.P1757 通天之分组背包2024-03-0412.P2135 方块消除2024-03-0613.D - String Bags2024-03-1014.P4395 [BOI2003] Gem 气垫车2024-03-1015.A - A Multiply2024-03-2416.P2642 双子序列最大和2024-03-2517.P1121 环状最大两段子段和2024-03-2518.P2422 良好的感觉2024-03-2519.P2854 [USACO06DEC] Cow Roller Coaster S2024-03-2620.C. Theofanis' Nightmare2024-03-2721.D. Birthday Gift2024-03-2722.P1284 三角形牧场2024-03-2823.P1353 [USACO08JAN] Running S2024-03-2824.P8736 [蓝桥杯 2020 国 B] 游园安排2024-03-3025.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0126.P7859 [COCI2015-2016#2] GEPPETTO2024-04-0527.P1435 [IOI2000] 回文字串2024-05-0828.P2340 [USACO03FALL] Cow Exhibition G2024-05-1329.P1854 花店橱窗布置2024-05-1330.P3147 [USACO16OPEN] 262144 P2024-05-1331.Game on Tree2024-05-1432.P2938 [USACO09FEB] Stock Market G2024-05-1833.P9691 [GDCPC2023] Base Station Construction2024-05-1834.P8624 [蓝桥杯 2015 省 AB] 垒骰子2024-05-2035.E. Money Buys Happiness2024-05-2236.P8675 [蓝桥杯 2018 国 B] 搭积木2024-05-2237.P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫2024-05-2238.P8806 [蓝桥杯 2022 国 B] 搬砖2024-05-2339.P5662 [CSP-J2019] 纪念品2024-05-2440.B. Mashmokh and ACM2024-06-0841.C. Minimizing the Sum2024-06-1242.H. Don't Blame Me2024-06-1343.D. Bicolorings2024-06-1444.D. Armchairs2024-06-1845.C. Chef Monocarp2024-06-1846.D. Invertible Bracket Sequences2024-06-2047.P1351 [NOIP2014 提高组] 联合权值2024-06-26

48.D. World is Mine2024-07-02

49.F. Feed Cats2024-07-0750.P3522 [POI2011] TEM-Temperature2024-07-2151.P3572 [POI2014] PTA-Little Bird2024-07-2252.P3089 [USACO13NOV] Pogo-Cow S2024-07-2253.E - Maximum Glutton2024-07-2854.Projects2024-07-2855.Elevator Rides2024-07-28

收起

原题链接

题解

1.alice的策略一定是从小到大一个一个拿
2.为了让alice拿不到某特定值的蛋糕，bob需要在alice拿它之前把它拿完
3.在最优策略中，bob一定可以从小拿到大
4.设此时bob要拿完第 $i$ 类蛋糕，该类蛋糕个数为 $k$ 则拿完这个蛋糕bob还有 $i-k-1$ 个回合可以用，所以从前面 $[1,i-k-1]$ 个回合里看看能挑完几个
5.每挑完一类蛋糕，挑后面蛋糕能用的回合就少一

code

#include<bits/stdc++.h>
using namespace std;
int rounds[5005]={0};
int dp[5005]={0};
int main()
{
    ios::sync_with_stdio(false);cout.tie(0);cin.tie(0);
    int t;
    cin>>t;

    while(t--)
    {
        int n;
        cin>>n;

        map<int,int> q;
        for(int i=1;i<=n;i++)
        {
            int x;
            cin>>x;
            q[x]++;
        }

        int cnt=0;
        for(auto it:q)
        {
            rounds[++cnt]=it.second;
        }

        int ans=0;
        for(int i=1;i<=cnt;i++)
        {
            if(rounds[i]<i)
            {
                int hh=rounds[i];
                //printf("di %d ge, it take %d rounds to clean\n",i,hh);
                for(int j=i-hh-1;j>=0;j--)
                {
                    if((i-dp[j]-1)<hh+j) continue;
                    dp[j+hh]=max(dp[j+hh],dp[j]+1);
                    ans=max(ans,dp[j+hh]);
                }
            }
        }


        for(int i=1;i<=n;i++)
        {
            //printf("huihe:%d  nadiao:%d\n",i,dp[i]);
            rounds[i]=0;
            dp[i]=0;
        }
        cout<<q.size()-ans<<'\n';
    }
    return 0;
}