#HDU 3371 Connect the Cities 【并查集、染色】

题目：

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15440 Accepted Submission(s): 4085

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

Sample Output

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

思路：题意为某地发洪水，某些地方还连在一起，有些地方已经淹掉了，需要将这些城市连在一起。

先将仍然连在一起的城市归入同一并查集，然后开始Kruskal算法即可。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string.h>
#include<stdio.h>

using namespace std;

struct data{
	short be, ed;
	short len;
}dat[25005];;
short bcj[510];
int ans, node;

#ifdef DEBUG

#endif

short find(int x)
{
	if (bcj[x] == 0)
	{
		bcj[x] = x;
	}
	if (bcj[x] == x)
	{
		return x;
	}
	bcj[x] = find(bcj[x]);
	return bcj[x];
}

void merge(int x, int y)
{
	int a = find(x);
	int b = find(y);
	bcj[a] = b;
	return;
}

bool cmp(data a, data b)
{
	return a.len < b.len;
}

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		memset(bcj, 0, sizeof(bcj));
		ans = 0;
		int n, m, k;
		cin >> n >> m >> k;
		for (size_t i = 0; i < m; i++)
		{
			scanf("%d%d%d", &dat[i].be, &dat[i].ed, &dat[i].len);
			//cin >> dat[i].be >> dat[i].ed >> dat[i].len;
		}
		sort(dat, dat + m, cmp);
		for (size_t i = 0; i < k; i++)
		{
			int temp_using, temp_tar, temp_numb;
			scanf("%d", &temp_numb);
			//cin >> temp_numb;
			if (temp_numb != 0)
			{
				scanf("%d", &temp_tar);
				//cin >> temp_tar;
				temp_tar = find(temp_tar);
			}
			for (size_t s = 1; s < temp_numb; s++)
			{
				scanf("%d", &temp_using);
				//cin >> temp_using;
				if (bcj[temp_using] == 0)
				{
					bcj[temp_using] = temp_tar;
				}
				else
				{
					merge(temp_using, temp_tar);
				}
			}
		}
		for (size_t i = 0; i < m; i++)
		{
			if (find(dat[i].be) != find(dat[i].ed))
			{
				merge(dat[i].be, dat[i].ed);
				ans += dat[i].len;
			}
		}
		int findans = find(1);
		for (size_t i = 1; i <= n; i++)
		{
			if (find(i) != findans)
			{
				printf("-1\n");
				findans = -1;
				break;
			}
		}
		if (findans != -1)
		{
			cout << ans << "\n";
		}
	}
}