'commit'

TangDou/AcWing_TiGao/T6/PaiXu/105.cpp

@@ -0,0 +1,52 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long LL;
+
+const int N = 100010;
+
+int row[N], col[N], s[N], c[N];
+
+LL solve(int n, int a[]) {
+    int sum = 0;
+    for (int i = 1; i <= n; i++) sum += a[i];
+    // 不能整除，最终无法完成平均工作
+    if (sum % n) return -1;
+    // 平均数
+    int avg = sum / n;
+
+    // 构建c数组
+    for (int i = 1; i <= n; i++) c[i] = c[i - 1] + a[i] - avg;
+    // 排序，为求中位数做准备
+    sort(c + 1, c + n + 1);
+    // 计算每个c[i]与中位数的差,注意下标从1开始时的写法 c[(n+1)/2]
+    LL res = 0;
+    for (int i = 1; i <= n; i++) res += abs(c[i] - c[(n + 1) / 2]);
+
+    return res;
+}
+
+int n, m, T; // n行,m列，对T个摊点感兴趣
+int main() {
+    // 加快读入
+    ios::sync_with_stdio(false), cin.tie(0);
+    cin >> n >> m >> T;
+    while (T--) {
+        int x, y;
+        cin >> x >> y;
+        row[x]++, col[y]++; // x行感兴趣的摊点数+1,y列感兴趣的摊点数+1
+    }
+
+    LL r = solve(n, row), c = solve(m, col);
+
+    if (~r && ~c)
+        printf("both %lld\n", r + c);
+    else if (~r)
+        printf("row %lld\n", r);
+    else if (~c)
+        printf("column %lld\n", c);
+    else
+        printf("impossible\n");
+
+    return 0;
+}