AcWing

https://www.acwing.com/problem/content/107/

很显然行和列是独立的，求出平均值之后变成“一个环，交换最小的次数使得各个位置相等”。假设要交换，则考虑最大的数，它要么把货物传递到左边，要么把货物传递到右边。多出来的部分一直累过去，假如多出来的部分是负的则会欠债，而且这个相差的值是会每次累计到总成本里面的。

这样做是错的，因为不能保证最大的数就一定是边界，也就是最大的数不一定不从其他地方接收牌。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int n, m, t;
int x[200005];
int y[200005];

ll calc_c() {
    int maxid = 0;
    for(int j = 1; j <= m; ++j) {
        if(y[j] > y[maxid])
            maxid = j;
    }
    int mean = t / m;
    ll sum = 0, cur = 0;
    for(int j = maxid, k = 1; k <= m; ++j, ++k) {
        cur += y[j] - mean;
        sum += abs(cur);
    }
    ll tsum = sum;
    sum = 0, cur = 0;
    for(int j = maxid + m, k = 1; k <= m; --j, ++k) {
        cur += y[j] - mean;
        sum += abs(cur);
    }
    return min(sum, tsum);
}

ll calc_r() {
    int maxid = 0;
    for(int i = 1; i <= n; ++i) {
        if(x[i] > x[maxid])
            maxid = i;
    }
    int mean = t / n;
    ll sum = 0, cur = 0;
    for(int i = maxid, k = 1; k <= n; ++i, ++k) {
        cur += x[i] - mean;
        sum += abs(cur);
    }
    ll tsum = sum;
    sum = 0, cur = 0;
    for(int i = maxid + n, k = 1; k <= n; --i, ++k) {
        cur += x[i] - mean;
        sum += abs(cur);
    }
    return min(sum, tsum);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d%d", &n, &m, &t);
    for(int ti = 1; ti <= t; ++ti) {
        int u, v;
        scanf("%d%d", &u, &v);
        x[u]++, x[u + n]++;
        y[v]++, y[v + m]++;
    }
    if((t % n) && (t % m))
        printf("impossible");
    else {
        ll sum = 0;
        if(t % n) {
            printf("column");
            sum = calc_c();
        } else if(t % m) {
            printf("row");
            sum = calc_r();
        } else {
            printf("both");
            sum = calc_r() + calc_c();
        }
        printf(" %lld
", sum);
    }

}

正确的做法把行列分开看，问题变成了“糖果传递”。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int n;
int a[2000005];
ll s[2000005];

ll calc() {
    ll sum = 0;
    for(int i = 1; i <= n; ++i)
        sum += a[i];
    int mean = sum / n;
    for(int i = 1; i <= n; ++i) {
        s[i] = s[i - 1] + a[i] - mean;
    }
    sort(s + 1, s + 1 + n);
    ll res = 0;
    int m = s[(n + 1) >> 1];
    for(int i = 1; i <= n; ++i)
        res += abs(s[i] - m);
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    printf("%lld
", calc());
}

改成这样：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int n, m, t;
int x[2000005];
int y[2000005];
ll s[2000005];

ll calc_r() {
    int mean = t / n;
    for(int i = 1; i <= n; ++i)
        s[i] = s[i - 1] + x[i] - mean;
    sort(s + 1, s + 1 + n);
    ll res = 0;
    ll mid = s[(n + 1) >> 1];
    for(int i = 1; i <= n; ++i)
        res += abs(s[i] - mid);
    return res;
}

ll calc_c() {
    int mean = t / m;
    for(int j = 1; j <= m; ++j)
        s[j] = s[j - 1] + y[j] - mean;
    sort(s + 1, s + 1 + m);
    ll res = 0;
    ll mid = s[(m + 1) >> 1];
    for(int j = 1; j <= m; ++j)
        res += abs(s[j] - mid);
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    scanf("%d%d%d", &n, &m, &t);
    for(int ti = 1; ti <= t; ++ti) {
        int u, v;
        scanf("%d%d", &u, &v);
        x[u]++, x[u + n]++;
        y[v]++, y[v + m]++;
    }
    if((t % n) && (t % m))
        printf("impossible");
    else {
        ll sum = 0;
        if(t % n) {
            printf("column");
            sum = calc_c();
        } else if(t % m) {
            printf("row");
            sum = calc_r();
        } else {
            printf("both");
            sum = calc_r() + calc_c();
        }
        printf(" %lld
", sum);
    }
}

PS：行和列写反都可以过5个点？？？