D. Yet Another Monster Fight

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-08

6.D. Yet Another Monster Fight2023-12-11

7.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

1.导论

这道题能不能用贪心做？答案是不能，具体为什么已经有题解给出回答。当贪心无法求解时，我们考虑一下动态规划。

2.算法设计

对于任一节点，其最坏情况（即所需最大起始威力值，后文称最大值）是什么？

当第一个被攻击的怪物（以下称头怪物）在其右边时，其最大值为右边怪物的数量加上自身初始值，头怪物在左边时同理。

因此我们可以考虑$o(n^2)$的做法，遍历一遍序列为头节点，再遍历一遍序列找出其余节点对应的最大值即可。

3.算法改进

遗憾的是，n最大达到了$3\cdot {10}^5$，因此我们要考虑优化程序。

我们发现，任一节点的最大值是由它和头怪物的位置关系决定的。

假设$a[i]$为头怪物，其左边节点的最大值为$r[j],j\in [1,i-1]$,右边节点的最大值为$l[j],j\in [i+1,n]$，因此我们可以正序倒序遍历一遍序列，求出每个点作为头怪物时左右节点的最大值。设$l1[i]$为左边节点的最大值,$r1[i]$为右边节点的最大值。

状态转移方程为：

$l1[i]=max(l1[i-1],r[i-1])$
$r1[i]=max(r1[i+1],l[i+1])$

4.代码呈现

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[300005]={0};//代表每个元素的初始值
ll l[300005]={0};//假如头怪物在左边的最坏情况，即左边所有怪物都清空了闪电才劈过来
ll r[300005]={0};
ll l1[300005]={0};//当前怪物为头怪物时，左边的最大值
ll r1[300005]={0};
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        l[i]=a[i]+i-1;
        r[i]=a[i]+n-i;
    }
    ll ans=3e9;
    for(int i=1;i<=n;i++) l1[i]=max(l1[i-1],r[i-1]);//由于头怪物在他们的右方，所以左边怪物的最大值等于把右边怪物都清空后的最坏情况
    for(int i=n;i>=1;i--) r1[i]=max(r1[i+1],l[i+1]);
    for(int i=1;i<=n;i++) ans=min(ans,max(max(l1[i],r1[i]),a[i]));//左边最大值，右边最大值，和自己
    cout<<ans;
    return 0;
}