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https://lydsy.com/JudgeOnline/problem.php?id=4883
answer
Each row and each column must be covered.
Consider for each row and each column to establish a point, and even between the row corresponding to one side is the right point coordinates.
In this way, each point must have a side to indicate that the row / column selection this side.
Each point must have a side of the base ring is a forest tree.
Therefore, with the smallest ring kruskal maintain direct forest trees.
The maintenance method is probably disjoint-set time and then maintains a communication information indicating that memory block is not present the ring. You can contribute an edge, the first time a communication is not communication in two blocks and two blocks do not have the communication ring; the second time is two communication endpoints in a block, and this block is not the communication loop.
Time complexity \ (O (nm \ log nm) \) .
#include<bits/stdc++.h>
#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}
typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;
template<typename I> inline void read(I &x) {
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}
const int N = 100000 + 7;
int n, m, tot;
int fa[N], hs[N];
struct Edges { int x, y, w; } e[N];
inline bool operator < (const Edges &a, const Edges &b) { return a.w < b.w; }
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline void work() {
std::sort(e + 1, e + n * m + 1);
ll ans = 0;
for (int i = 1; i <= n + m; ++i) fa[i] = i;
for (int i = 1; i <= n * m; ++i) {
int x = find(e[i].x), y = find(e[i].y), z = e[i].w;
// dbg("z = %d\n", z);
if (x == y) {
if (!hs[x]) hs[x] = 1, ans += z;//, dbg("****** %d %d %d\n", e[i].x, e[i].y, z);
continue;
}
if (hs[x] && hs[y]) continue;
ans += z, fa[y] = x, hs[x] |= hs[y];//, dbg("****** %d %d %d\n", e[i].x, e[i].y, z);
}
printf("%lld\n", ans);
}
inline void init() {
read(n), read(m);
int v;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) {
read(v);
e[++tot] = (Edges){ i, j + n, v };
}
}
int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif
init();
work();
fclose(stdin), fclose(stdout);
return 0;
}