hdu 1233 还是畅通工程 并查集or最小生成树

某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。 

Input测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。 
当N为0时，输入结束，该用例不被处理。 
Output对每个测试用例，在1行里输出最小的公路总长度。 
Sample Input

3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0

Sample Output

3
5


        
 
Huge input, scanf is recommended.

• 并查集的

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<map>
using namespace std;
int n,bc[110],ans;
struct P {
	int a, b, c;
	bool operator<(const P&a)const {
		return c < a.c;
	}
}s[5005];
int Find(int i) {
	if (bc[i] == i)
		return i;
	else {
		bc[i] = Find(bc[i]);
		return bc[i];
	}
}
void Join(int a, int b, int idx) {
	int x = Find(a), y = Find(b);
	if (x != y) {
		bc[y] = x;
		ans += s[idx].c;
	}
}
int main() {
	//freopen("in.txt", "r", stdin);
	while (scanf("%d",&n)&&n)
	{
		int m = (n*(n - 1)) / 2;
		ans = 0;
		for (int i = 1; i <= n; i++)
			bc[i] = i;
		for (int i = 0; i < m; i++) {
			scanf("%d %d %d", &s[i].a, &s[i].b, &s[i].c);
		}
		sort(s, s + m);
		for (int i = 0; i < m; i++)
			Join(s[i].a, s[i].b, i);
		printf("%d
", ans);
	}
	return 0;
}

• 　　最小生成树

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<map>
using namespace std;
#define inf 0x3f3f3f
int n,ans;
int edge[110],M[110][110];
int main() {
	//freopen("in.txt", "r", stdin);
	while (scanf("%d",&n)&&n)
	{
		memset(M, inf, sizeof(M));
		int m = (n*(n - 1)) / 2;
		ans = 0;
		for (int i = 0; i < m; i++) {
			int a, b, c;
			scanf("%d %d %d", &a, &b, &c);
			M[a][b] = M[b][a] = c;
		}
		bool S[110];
		memset(S, 0, sizeof(S));
		for (int i = 2; i <= n; i++)
			edge[i] = M[1][i];
		S[1] = 1;
		for (int i = 1; i <= n-1; i++) {
			int v=-1, maxl = inf;
			for (int j = 2; j <= n; j++)
				if (!S[j]&&maxl > edge[j]) {
					v = j;
					maxl = edge[j];
				}
			if (v == -1) break;
			S[v] = 1;
			ans += edge[v];
			for (int j = 2; j <= n; j++)
				if (!S[j] && M[v][j] < edge[j]) {
					edge[j] = M[v][j];
				}
		}
		printf("%d
", ans);
	}
	return 0;
}