F. Feed Cats

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2024-06-2648.D. World is Mine2024-07-02

49.F. Feed Cats2024-07-07

50.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

每个点要么喂，要么不喂，我们令 $dp[i]$ 为前 $i$ 个步骤最多能喂养多少猫，易得 $dp[i]$ 是单调不减的
我们再维护每个点被包含的区间里的最左端 $l$
这样一来 $dp[i]=max(dp[i-1],dp[l-1]+sum)$

可是如何维护每个点被包含区间的最左端呢？
我们先记录下每个右端点的最小左端点，然后逆序遍历，即对于 $i$ 求以 $[i,n]$ 内为右端点的最小左端点

code

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

int farl[2000005],pre[1000006]={0},dp[1000006];
int main()
{
    int t;
    cin>>t;

    while(t--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            farl[i]=n+1;
            pre[i]=0;
        }

        for(int i=1;i<=m;i++)
        {
            int x,y;
            cin>>x>>y;
            pre[x]++;
            pre[y+1]--;

            farl[y]=min(farl[y],x);
        }

        for(int i=n-1;i>=1;i--) farl[i]=min(farl[i],farl[i+1]);

        int sum=0,ans=0;
        for(int i=1;i<=n;i++)
        {
            sum+=pre[i];

            dp[i]=dp[i-1];
            if(farl[i]<=i) dp[i]=max(dp[i],dp[farl[i]-1]+sum);
            ans=max(ans,dp[i]);
        }
        cout<<ans<<'\n';
    }
    return 0;
}