AtCoder AGC030B Tree Burning

Posted 2022-07-19 suncongbo

题目链接

https://atcoder.jp/contests/agc030/tasks/agc030_b

题解

细节好题。。
首先假设第一步往右走，那么可以发现当拐弯的次数一定时路径是唯一的
于是可以枚举这个值
然后很烦的是枚举之后要分奇偶讨论。。
最后再翻过来做一遍处理第一步往左走就行了
时间复杂度\(O(n)\)

代码

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define llong long long
using namespace std;

void read(int &x)

    int f=1;x=0;char s=getchar();
    while(s<'0'||s>'9')if(s=='-')f=-1;s=getchar();
    while(s>='0'&&s<='9')x=x*10+s-'0';s=getchar();
    x*=f;


const int N = 2e5;
llong a[N+3];
llong s[N+3];
int n; llong m,ans;

void solve()

    s[0] = 0ll; for(int i=1; i<=n; i++) s[i] = s[i-1]+a[i];
    for(int i=1; i<=n; i++)
    
        llong tmp;
        if(i&1)
        
            int x = (i-1)>>1;
            tmp = (m*x-(s[n]-s[n-x]))*2+a[n-x]+(s[n-x-1]-s[n-x-x-1])*2;
        
        else
        
            int x = i>>1;
            tmp = (m*(x-1)-(s[n]-s[n-x+1]))*2+(m-a[n-x+1])+(s[n-x]-s[n-x-x])*2;
        
        ans = max(ans,tmp);
    


int main()

    scanf("%lld%d",&m,&n);
    for(int i=1; i<=n; i++) scanf("%lld",&a[i]);
    solve();
    for(int i=1; i<=n; i++) a[i] = m-a[i];
    reverse(a+1,a+n+1);
    solve();
    printf("%lld\n",ans);
    return 0;