codeforces248(div1) A. Ryouko's Memory Note

标签：codeforces   链表   

可以把序列中一个数改变，使得序列中后一个数与前一个数差的绝对值之和最小

把与数x相邻的数加入G[x]的链表中(若这个数值也为x，则不加入)   那么改变了数x，则相当于只会影响到这个链表中的数

为了让x变化后的数  与这个链表中的数差值绝对值之和最小   取排序后的序列的中位数即可

//#pragma comment(linker, "/STACK:102400000,102400000")
//HEAD
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>

#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstdlib>
#include <list>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//STL
#define PB push_back
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)


#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
#define SZ(V) (int)V.size()
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
#define sqr(x) x * x

typedef vector <int> VI;
typedef unsigned long long ULL;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 100010;
const double eps = 1e-10;
const LL MOD = 1e9 + 9;

int a[maxn];
VI G[maxn];

int main()
{
    int n, m;
    while (~RII(n, m))
    {
        REP(i, n + 1)   G[i].clear();
        FE(i, 1, m)
           RI(a[i]);
        LL tot = 0;
        FE(i, 1, m)
        {
            if (i != 1 && a[i] != a[i - 1])
                G[a[i]].push_back(a[i - 1]);
            if (i != m && a[i] != a[i + 1])
                G[a[i]].push_back(a[i + 1]), tot += abs(a[i + 1] - a[i]);
        }
        LL ans = tot;
//        cout << tot << "----" <<endl;
        FE(i, 1, n)
        {
            if (G[i].size() == 0) continue;
            LL t = tot;
//            cout << t1 << "  ----x:";
            sort(G[i].begin(), G[i].end());
            int x = G[i][G[i].size() / 2];
            REP(j, G[i].size())
                t += abs(G[i][j] - x) - abs(G[i][j] - i);
            ans = min(ans, t);
//            cout << x <<"   t2: " << t2 <<endl;
        }
//        if (m == 1) ans = 0;
        cout << ans << endl;
    }
    return 0;
}
/*
4 6
1 2 3 4 3 2

10 5
9 4 3 8 8

1000 5
0 1 1000 999 1
*/

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codeforces248(div1) A. Ryouko's Memory Note

标签：codeforces   链表   

原文地址：http://blog.csdn.net/colin_27/article/details/37907129