D. Slimes

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 Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.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 B2024-02-28

52.D. Slimes2024-02-28

53.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

收起

原题链接

题解

对于任何一个粘液块s而言，要么是从左边被吞并，要么是从右边被吞并，根据对称性，两边的决策是一样的，因此先考虑右边

对于被右边吞并而言，有以下几个特征
1.起始粘液一定是吞掉了s右边一整块连续的粘液
2.右边区间一定存在大小不同的相邻粘液，这样才能发动吞并
3.由一二猜想，只要存在不同的相邻粘液，这一片区间上的粘液都会被吞并至只剩下一个粘液
证明：
假如只存在一个不同相邻粘液
最大的吞并比他小的-->原来大的比同伴大-->吞并同伴
存在两个及以上同理
4.因此我猜想，对于右边区间而言，存在不同相邻粘液块且区间和大于s的最小区间就是答案
开始证明
运用反证法：如果该区间不是最小区间，那么区间缩小，区间和会减小以至于可能小于s，也可能不存在相同相邻粘液，与题意不符，得证

然后很容易发现区间和具有单调性，是否存在不同相邻粘液也有单调性（不会减小）

因此二分

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll pres[300005]={0};
ll sufs[300005]={0};
ll a[300005]={0};
ll prech[300005]={0};
ll sufch[300005]={0};
ll press(ll l,ll r,ll x)
{
    r++;
    l--;
    ll i=l;
    if(a[i]<a[i+1])return i+1;
    while(l+1<r)
    {
        ll mid=(l+r)/2;
        if(pres[mid]-pres[i]>a[i]&&prech[mid]-prech[i+1]>0)
            r=mid;
        else l=mid;
    }
    return r;
}

ll sufss(ll l,ll r,ll x)
{
    l--;
    r++;
    ll i=r;
    if(a[i-1]>a[i])return i-1;
    while(l+1<r)
    {
        ll mid=(l+r)/2;
        if(sufs[mid]-sufs[i]>a[i]&&sufch[mid]-sufch[i-1]>0)
            l=mid;
        else r=mid;
    }
    return l;
}
int main()
{
    ll t;
    cin>>t;
    while(t--)
    {
        ll n;
        cin>>n;
        for(ll i=1;i<=n;i++)
        {
            cin>>a[i];
            pres[i]=pres[i-1]+a[i];
            if(i!=1)prech[i]=prech[i-1]+(a[i]!=a[i-1]);
        }

        sufs[n]=a[n];
        sufch[n]=0;
        for(ll i=n-1;i>=1;i--)
        {
            sufs[i]=sufs[i+1]+a[i];
            sufch[i]=sufch[i+1]+(a[i]!=a[i+1]);
        }

        for(ll i=1;i<=n;i++)
        {
            if(i==1)
            {
                ll r=press(2,n,a[1]);
                if(r!=n+1) cout<<r-i<<" ";
                else printf("-1 ");
            }
            else if(i==n)
            {
                ll l=sufss(1,n-1,a[1]);
                if(l!=0) cout<<i-l<<" ";
                else printf("-1 ");
            }
            else
            {
                ll r=press(i+1,n,a[i]);
                ll l=sufss(1,i-1,a[i]);
                if(l!=0&&r!=n+1)cout<<min(i-l,r-i)<<" ";
                else if(l!=0) cout<<i-l<<" ";
                else if(r!=n+1) cout<<r-i<<" ";
                else printf("-1 ");
            }
        }

        puts("");
    }
    return 0;
}