《纸牌问题》

贪心的问题总是很难证明。

对于单向传递的纸牌问题：维护差值的前缀和。

https://www.luogu.com.cn/problem/P1031

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e6+5;
const int M = 1e7+5;
const LL Mod = 20101009;
#define pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int a[N],sum = 0;
int main()
{
    int n;n = read();
    for(int i = 1;i <= n;++i) a[i] = read(),sum += a[i];
    sum /= n;
    for(int i = 1;i <= n;++i) a[i] = a[i] - sum;
    int ans = 0;
    for(int i = 1;i <= n;++i)
    {
        if(a[i] == 0) continue;
        a[i + 1] += a[i];
        ans++;
    }
    printf("%d
",ans);
    system("pause");
    return 0;
}

https://www.luogu.com.cn/problem/P2512

当这个问题变成环形之后，虽然可以两向传递，但是按照差值维护的思想，我们只需要传递负的价值即可。

所以就是一个单向的环问题，那么对于任意一处k断开链。

那么对于k处断开后，i位置的代价就是abs(s[i] - s[k])

那么$ans = sum_{i=1}^{N}|S[i]-S[k]$，显然当k取s的中位数时最优。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e6+5;
const int M = 1e7+5;
const LL Mod = 20101009;
#define pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int a[N];
LL sum = 0;
int main()
{
    int n;n = read();
    for(int i = 1;i <= n;++i) a[i] = read(),sum += a[i];
    sum /= n;
    for(int i = 1;i <= n;++i) a[i] = a[i] - sum;
    for(int i = 1;i <= n;++i) a[i + 1] += a[i];
    sort(a + 1,a + n + 1);
    LL ans = 0;
    if(n % 2 != 0)
    {
        int mid = n / 2 + 1;
        for(int i = 1;i <= n;++i) ans += abs(a[i] - a[mid]);
    }
    else
    {
        int mid1 = n / 2,mid2 = n / 2 + 1;
        LL ans1 = 0,ans2 = 0;
        for(int i = 1;i <= n;++i) ans1 += abs(a[i] - a[mid1]);
        for(int i = 1;i <= n;++i) ans2 += abs(a[i] - a[mid2]);
        ans = min(ans1,ans2);
    }
    printf("%lld
",ans);
    system("pause");
    return 0;
}