【次小生成树】POJ 1679 The Unique MST

http://poj.org/problem?id=1679


题意：问最小生成树是否唯一

思路：
用Kruskal先求最小生成树，结果即为min，把所用到的边记录下来（这里是记录的对应的下表），然后枚举这些边，
每次去掉一个边再求一次最小生成树，结果为tmin，如果能构成最小生成树tmin==min，则说明最小生成树不唯一

Sample Input
2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output
3
Not Unique!

代码自己YY吧……

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <set>
//#include <map>
#include <queue>
#include <utility>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
//#include <ctime>
#include <ctype.h>
using namespace std;
#define inf 0x3fffffff

int n, k, pre[105], num[10005], ind;
struct road{
	int a, b;
	int weight;
}r[10005];

bool cmp (road x, road y)
{
	return x.weight < y.weight;
}

int find (int a)
{
	while (a != pre[a])
		a = pre[a];
	return a;
}

int MST (int key)
{
	int mincost = 0, i, tp = 0, A, B;
	for (i = 1; i <= n; i++)
		pre[i] = i;
	for (i = 0; i < k; i++)
	{
		if (i == key)
			continue;
		A = find (r[i].a);
		B = find (r[i].b);
		if (A != B)
		{
			if (key == -1)
				num[ind++] = i;
			mincost += r[i].weight;
			pre[B] = A;
		}
	}
	for (i = 1; i <= n; i++)
	{
		if (pre[i] == i)
		{
			tp++;
			if (tp > 1)
				return inf;
		}
	}
	return mincost;
}

int main()
{
	int m, t, i, a, b, w, mins, tmins, tp;
	scanf ("%d", &t);
	while (t--)
	{
		k = ind = 0;
		scanf ("%d%d", &n, &m);
		while (m--)
		{
			scanf ("%d%d%d", &a, &b, &w);
			r[k].a = a;
			r[k].b = b;
			r[k].weight = w;
			k++;
		}
		sort (r, r+k, cmp);
		mins = MST (-1);
		tmins = inf;
		for (i = 0; i < ind; i++)
		{
			tp = MST (num[i]);
			if (tmins > tp)
				tmins = tp;
		}
		if (tmins > mins)
			printf ("%d\n", mins);
		else puts ("Not Unique!");
	}
	return 0;
}