P3089 [USACO13NOV] Pogo-Cow S

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 Mine2024-07-0249.F. Feed Cats2024-07-0750.P3522 [POI2011] TEM-Temperature2024-07-2151.P3572 [POI2014] PTA-Little Bird2024-07-22

52.P3089 [USACO13NOV] Pogo-Cow S2024-07-22

53.E - Maximum Glutton2024-07-2854.Projects2024-07-2855.Elevator Rides2024-07-28

收起

原题链接

题解

暴力dp：
遍历 $i,j,k$ ，$dp[i][j]=max(dp[j][k])+v_i$ 其中 $x_i-x_j\ge x_j-x_k$

优化：
对于 $j$ 来说 ，随着 $i$ 越大， $k$ 可以越小，因此省去了遍历一层 $k$ ，而是维护每个点的 $k$ ，（反正求的是最大值）

细节

1.有两个方向
2.任意起点

code

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

struct node
{
    int x,v;
}star[1005];

int mx[1005]={0},last[1005];
int dp[1005][1005]={0};

bool cmp1(node a,node b)
{
    return a.x<b.x;
}
bool cmp2(node a,node b)
{
    return a.x>b.x;
}


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

    int ans=0;
    for(int i=1;i<=n;i++)
    {
        cin>>star[i].x>>star[i].v;
        ans=max(ans,star[i].v);
    }

    sort(star+1,star+1+n,cmp1);
    for(int i=1;i<=n;i++)
    {
        last[i]=i-1;
        mx[i]=star[i].v;
        for(int j=1;j<i;j++)
        {
            while(last[j]>0&&star[i].x-star[j].x>=star[j].x-star[last[j]].x)
            {
                mx[j]=max(mx[j],dp[j][last[j]]);
                last[j]--;
            }
            dp[i][j]=mx[j]+star[i].v;
            ans=max(ans,dp[i][j]);
            //printf("i:%d   j:%d  dp:%d\n",i,j,dp[i][j]);
        }
        dp[i][i]=star[i].v;
       // printf("i:%d   j:%d  dp:%d\n",i,i,dp[i][i]);
    }


    sort(star+1,star+1+n,cmp2);

    for(int i=1;i<=n;i++)
    {
        last[i]=i-1;
        mx[i]=star[i].v;
        for(int j=1;j<i;j++)
        {
            while(last[j]>0&&star[j].x-star[i].x>=star[last[j]].x-star[j].x)
            {
                mx[j]=max(mx[j],dp[j][last[j]]);
                last[j]--;
            }
            dp[i][j]=mx[j]+star[i].v;
            ans=max(ans,dp[i][j]);
          //  printf("i:%d   j:%d  dp:%d\n",i,j,dp[i][j]);
        }
        dp[i][i]=star[i].v;
       // printf("i:%d   j:%d  dp:%d\n",i,i,dp[i][i]);
    }

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