poj_3371Connect the Cities（最小生成树）

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5793    Accepted Submission(s): 1701

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  

 

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

 

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

 

Sample Input

1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

 

Sample Output

1

开始判断sum值，wa了，看题目(0 <= c <= 1000)，并不能从sum=0判断的=。=

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#pragma warning(disable : 4996)
#define MAX 25005
typedef struct edge
{
	int x, y;
	int w;
}edge;

edge e[MAX];
int father[MAX], ranks[MAX];

bool cmp(edge a,edge b)
{
	return a.w < b.w;
}

void Make_Set(int n)
{
	for(int i = 1; i <= n; i++)
	{
		father[i] = i;
		ranks[i] = 0;
	}
}

int Find_Set(int x)
{
	if(x != father[x])
		father[x] = Find_Set(father[x]);
	return father[x];
}

void Merge_Set(int x, int y)
{
	x = Find_Set(x);
	y = Find_Set(y);
	if(x == y) return;
	if(ranks[x] > ranks[y])
	{
		father[y] = x;
	}
	else if(ranks[x] < ranks[y])
	{
		father[x] = y;
	}
	else 
	{
		ranks[y]++;
		father[x] = y;
	}
}

int main()
{
	freopen("in.txt", "r", stdin);
	int t, i, j, n, m, k, sum, x, y, w, v;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d %d %d", &n, &m, &k);
		Make_Set(n);
		for(i = 0; i < m; i++)
		{
			scanf("%d %d %d", &x, &y, &w);
			e[i].x = x;
			e[i].y = y;
			e[i].w = w;
		}

		sort(e, e + m, cmp);

		for(i = 1; i <= k; i++)
		{
			scanf("%d", &v);
			if(v == 2)
			{
				cin >> x >> y;
				Merge_Set(x, y);
			}
			else
			{
				scanf("%d %d", &x, &y);
				Merge_Set(x, y);
				x = y;
				for(j = 2; j < v; j++)
				{
					scanf("%d", &y);
					Merge_Set(x, y);
					x = y;
				}
			}
		}
		sum = 0;
		for(i = 0; i < m; i++)
		{
			x = Find_Set(e[i].x);
			y = Find_Set(e[i].y);
			if(x != y)
			{
				Merge_Set(x, y);
				sum += e[i].w;
			}
		}
		int count = 0;
		for(i = 1; i <= n; i++)
		{
			if(i == father[i])
			{
				count++;
			}
		}
		if(count > 1)
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n", sum);
		}
	}
	return 0;
}