qoj5371 Matrix （二分图匹配）

qoj5371 Matrix

二分图匹配

判断无解的情况，当且仅当有 \(a_{i,j}\) 为负数或每一行和每一列的和不相同时无解。

因为 \(m\le N^2\)，所以我们只需要每一次至少完成一个 \(a_{i,j}\) 即可。观察 \(B\) 矩阵的形成，实际上就是一个 \(i\) 行只能和一个 \(j\) 列匹配，跑二分图匹配即可。每一次求出 \(B\) 矩阵让其系数为其中所有 \(1\) 位置对应的 \(a_{i,j}\) 的最小值，那么对应的 \(a_{i,j}\) 减去系数后会若干位置会变成 \(0\)，此时边 \((i,j)\) 无法选择，退掉。重复上面的操作知道二分图最大匹配 \(\ne n\)，结束。

#include <bits/stdc++.h>
#define pii std::pair<int, int>
#define fi first
#define se second
#define pb push_back

using i64 = long long;
using ull = unsigned long long;
const i64 iinf = 0x3f3f3f3f, linf = 0x3f3f3f3f3f3f3f3f;
const int N = 60;

int n;
int a[N][N], vis[N], mat[N], sm1[N], sm2[N], p[N];

bool dfs(int u) {
	for (int i = 1; i <= n; i++) {
		if(a[u][i] && !vis[i]) {
			vis[i] = 1;
			if(!mat[i] || dfs(mat[i])) {
				mat[i] = u, p[u] = i;
				return 1;
			}
		}
	}
	return 0;
}
std::vector<std::vector<int> > ans;
void solve() {
	memset(sm1, 0, sizeof(sm1));
	memset(sm2, 0, sizeof(sm2));

	std::cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			std::cin >> a[i][j];
			sm1[i] += a[i][j];
			sm2[j] += a[i][j];
		}
	}

	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (a[i][j] < 0) {
		std::cout << "-1\n";
		return;
	}

	for (int i = 1; i <= n; i++) {                     
		if(sm1[i] != sm1[1] || sm2[i] != sm2[1]) {         
			std::cout << "-1\n";
			return;
		}
	}

	for (int i = 1; i <= n; i++) {
		memset(vis, 0, sizeof(vis));
		dfs(i);
	}

	bool ok = 1;
	while(ok) {
		int mn = iinf;
		for (int i = 1; i <= n; i++) {
			mn = std::min(mn, a[i][p[i]]);
		}
		std::vector<int> v;
		v.pb(mn);
		for (int i = 1; i <= n; i++) {
			a[i][p[i]] -= mn;
			v.pb(p[i]);
		}
		ans.pb(v);
		for (int i = 1; i <= n; i++) {
			if(!a[i][p[i]]) {
				mat[p[i]] = 0, p[i] = 0;
				memset(vis, 0, sizeof(vis));
				if(!dfs(i)) ok = 0; //判断是否仍有增广路
			}
		}
	}

	std::cout << ans.size() << "\n";
	for (int i = 0; i < ans.size(); i++) {
		for (auto x : ans[i]) {
			std::cout << x << " ";
		}
		std::cout << "\n";
	}	
	ans.clear();
}
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
	int t;
	std::cin >> t;

	while (t--) solve();

	return 0;
}