P2854 [USACO06DEC] Cow Roller Coaster 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2024-03-25

19.P2854 [USACO06DEC] Cow Roller Coaster S2024-03-26

20.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 Bird2024-07-2252.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.当没有花费限制的时候，我们可以将其抽象为简单的背包问题
2.如果有了花费限制，那么我们就再加一维条件

3.如果一个线段能用，那么它前面一定是铺满的，那我们令线段按起点排序，通过某种运算，保证放这个线段时，前面的线段组成是最优的
比如在 $i$ 点结尾位置花费 $j$ 所达到的最优值，这个最优值与后面线段怎么摆无关，所以可以存起来减少重复运算，这也是dp的核心

code

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int start,len,v,c;
}seg[10005];
bool cmp(node a,node b)
{
    if(a.start!=b.start) return a.start<b.start;
    return a.len<b.len;
}

int f[1005][1005]={0};
int main()
{
    int l,n,b;
    cin>>l>>n>>b;
    for(int i=1;i<=n;i++)
    {
        cin>>seg[i].start>>seg[i].len>>seg[i].v>>seg[i].c;
    }

    sort(seg+1,seg+1+n,cmp);

    for(int i=0;i<=l;i++)
    {
        for(int j=1;j<=n;j++)
        {
            int start=seg[j].start;
            if(start==i)
            {
                int ends=start+seg[j].len;
                if(i==0) f[ends][seg[j].c]=max(seg[j].v,f[ends][seg[j].c]);
                for(int k=seg[j].c;k<=b;k++) if(f[start][k-seg[j].c]) f[ends][k]=max(f[start][k-seg[j].c]+seg[j].v,f[ends][k]);
            }
        }
    }

    int ans=0;
    for(int i=1;i<=b;i++) ans=max(ans,f[l][i]);
    cout<<(ans?ans:-1);
    return 0;
}