【Foreign】Bumb [模拟退火]

Bumb

Time Limit: 20 Sec  Memory Limit: 512 MB

Description

　　

Input

　　

Output

　　

Sample Input

　　4
　　1 5 1 4

Sample Output

　　5

HINT

　　

Solution

　　首先，我们对于一个已知的k，可以O(n)得到Ans，这样就有60%了。

　　那么怎么做90%呢？老老实实写O(nlogn)是不可能的！模拟退火美滋滋！

　　传授一点人生的经验吧：写个对拍用于调参，由于这种题不能很单峰，显然温度变化设大一点比较容易正确，无限running + 卡时即可。

　　是的，没有错！BearChild就这样拿到了90%！

　　我们来考虑100%怎么做，显然复杂度O(n)。我们把( a[i - 1], a[i] ]看作若干个区间挂在数轴上。

　　然后考虑k：1->m对于答案的变化。显然可以记录cnt表示k包含的区间个数，以及sum和，这样就可以做完啦！

Code

#include<iostream>    
#include<string>    
#include<algorithm>    
#include<cstdio>    
#include<cstring>    
#include<cstdlib>
#include<cmath>
#include<vector>
#include<ctime>
using namespace std;  
typedef long long s64;

const int ONE = 1000005;
const s64 INF = 1e18;

int n, m;
int a[ONE];
int Now, A;
s64 Ans = INF;

int get()
{
        int res=1,Q=1;char c;
        while( (c=getchar())<48 || c>57 ) 
        if(c=='-')Q=-1; 
        res=c-48;     
        while( (c=getchar())>=48 && c<=57 )
        res=res*10+c-48;    
        return res*Q;
}

void TimeE()
{
        if((double)clock() / CLOCKS_PER_SEC > 0.97)
        {
            printf("%lld", Ans);
            exit(0);
        }
}

s64 Get(int x, int y)
{
        TimeE();
        if(x <= y) return y - x;
        return y + m - x;
}

s64 Judge(int k)
{
        s64 res = 0;
        for(int i = 2; i <= n; i++)
            res += min(Get(a[i - 1], a[i]), 1 + Get(k, a[i]));
        Ans = min(Ans, res);
        return res;
}

double Random() {return rand() / (double)RAND_MAX;}
void SA(double T)
{
        Now = m / 2;
        while(T >= 1)
        {
            A = Now + (int)(T * (Random() * 2 - 1));
            if(A < 1 || A > m) A = T * Random();
            s64 dE = Judge(A) - Judge(Now);
            if(dE < 0) Now = A;
            T *= 0.92;
        }
}

int main()
{
        n = get();    m = get();
        for(int i = 1; i <= n; i++)
            a[i] = get();
        if(n > 3000) {for(;;)SA(m); return 0;}
        
        for(int i = 1; i <= m; i++)
            Ans = min(Ans, Judge(i));
        
        printf("%lld", Ans);
}

#include<iostream>    
#include<string>    
#include<algorithm>    
#include<cstdio>    
#include<cstring>    
#include<cstdlib>
#include<cmath>
#include<vector>
using namespace std;  
typedef long long s64;

const int ONE = 1000005;
const s64 INF = 1e18;

int n, m;
int a[ONE];
int L[ONE];
vector <int> R[ONE];
s64 Ans;

int get()
{
        int res=1,Q=1;char c;
        while( (c=getchar())<48 || c>57 ) 
        if(c=='-')Q=-1; 
        res=c-48;     
        while( (c=getchar())>=48 && c<=57 )
        res=res*10+c-48;    
        return res*Q;
}

int Get(int x, int y)
{
        if(x <= y) return y - x;
        return y + m - x;
}

int main()
{
        n = get();    m = get();
        for(int i = 1; i <= n; i++)
            a[i] = get();
        
        for(int i = 2; i <= n; i++)
        {
            Ans += Get(a[i - 1], a[i]);
            L[a[i - 1]]++, R[a[i]].push_back(a[i - 1]);
        }
        
        s64 sum = Ans, cnt = 0;
        
        for(int i = 2; i <= n; i++)
            if(a[i - 1] > a[i])
                sum -= m - a[i - 1], cnt++;
        
        for(int k = 2; k <= m; k++)
        {
            int len = R[k - 1].size();
            cnt = cnt + L[k - 2] - len;
            sum -= cnt;
            for(int i = 0; i < len; i++)
                sum += Get(R[k - 1][i], k - 1) - 1;
            Ans = min(Ans, sum);
        }
        
        printf("%lld", Ans);
}