POJ 1258 Agri-Net (Prim&Kruskal)

题意：FJ想连接光纤在各个农场以便网络普及，现给出一些连接关系（给出邻接矩阵），从中选出部分边，使得整个图连通。求边的最小总花费。

思路：裸的最小生成树，本题为稠密图，Prim算法求最小生成树更优，复杂度O(n^2)

prim：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int mat[110][110];
bool vis[110];
int d[110];

int Prim(int n) {
	int ans = 0;
	int p = 0;
	vis[0] = 1;
	for (int j = 1; j < n; ++j) {
		d[j] = mat[p][j];
	}
	for (int i = 1; i < n; ++i) {
		p = -1;
		for (int j = 1; j < n; ++j) {
			if (vis[j]) continue;
			if (p == -1 || d[j] < d[p]) {
				p = j;
			}
		}
		ans += d[p];
		vis[p] = 1;
		for (int j = 1; j < n; ++j) {
			if (vis[j]) continue;
			d[j] = min(d[j], mat[p][j]);
		}
	}
	return ans;
}

int main() {
	int n, i, j;
	while (scanf("%d", &n) != EOF) {
		memset(vis, 0, sizeof(vis));
		for (i = 0; i < n; ++i) {
			for (j = 0; j < n; ++j) {
				scanf("%d", &mat[i][j]);
			}
		}
		int ans = Prim(n);
		printf("%d\n", ans);
	}
	return 0;
}

kruskal：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int N; // 节点数量
struct edge {
	int from, to, dist;
	bool operator<(const edge &b) const {
		return dist < b.dist;
	}
} es[10006];
int par[105];
void init() {
	for (int i = 1; i <= N; ++i) par[i] = i;
}
int find(int x) {
	return x == par[x] ? x : par[x] = find(par[x]);
}
void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if (x != y) par[x] = y;
}
int kruskal() {
	int res = 0;
	init();
	int E = N*N;
	sort(es + 1, es + 1 + E);
	for (int i = 1; i <= E; ++i) {
		edge e = es[i];
		//printf("u:%d v:%d d:%d\n", e.from, e.to, e.dist);
		if (find(e.from) != find(e.to)) {
			unite(e.from, e.to);
			res += e.dist;
		}
	}
	return res;
}
void solve() {
	cout << kruskal() << endl;
}
int main()
{
	while (cin >> N) {
		int d;
		int id;
		for (int u = 1; u <= N; ++u)
		 for (int v = 1; v <= N; ++v) {
			cin >> d;
			id = (u - 1)*N + v;
			es[id].from = u;
			es[id].to = v;
			es[id].dist = d;
		}
		solve();
	}
	return 0;
}