Codeforces 959 B. Mahmoud and Ehab and the message(并查集)


题目链接

B. Mahmoud and Ehab and the message
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Mahmoud wants to send a message to his friend Ehab. Their language consists of n words numbered from 1 to n. Some words have the same meaning so there are k groups of words such that all the words in some group have the same meaning.

Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?

The cost of sending the message is the sum of the costs of sending every word in it.

Input

The first line of input contains integers nk and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.

The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.

The third line contains n integers a1a2...an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word.

The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.

The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.

Output

The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.

Examples
input
Copy
5 4 4
i loser am the second
100 1 1 5 10
1 1
1 3
2 2 5
1 4
i am the second
output
107
input
Copy
5 4 4
i loser am the second
100 20 1 5 10
1 1
1 3
2 2 5
1 4
i am the second
output
116
Note

In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.

In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.


题目大意:Mahmoud要发短信,每个单词有他的权值,并且单词之间如果是一组的就可以用同一组的代替。

问发送短信最小权值为多少。

因为数据量大,不能暴力。

对于同一组的单词我们可以用并查集优化查找过程,只记录队头的权值为最小权值。

然后对于每个单词hash就行了,这里hash用map吧。

有一点要注意的是答案会超int吧。

代码如下

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = (int)(1e5) + 100;
const int inf = 0x3f3f3f3f;
const int mod = 2520;
const double eps = 1e-3;
typedef long long LL;
typedef unsigned long long ull;
mapmp;
int val[maxn], gval[maxn];
struct unionfind {//启发式合并小树合到大树里,速度更优秀,但是队头编号不一定最小
	int pre[maxn], ranks[maxn];
	int find(int x) {
		if (pre[x] == x) return x;
		else return pre[x] = find(pre[x]);//压缩路径
	}
	void unite(int a, int b) {
		int x = find(a), y = find(b);
		if (x == y) return;
		if (ranks[x] < ranks[y]) pre[x] = y;//rank优化
		else {
			pre[y] = x;
			if (ranks[x] == ranks[y]) ranks[x]++;
		}
		return;
	}
	void init(int n) {
		for (int i = 0; i <= n; i++) {
			pre[i] = i;
			ranks[i] = 0;
		}
	}
}un;
char str[25];
int main() {
	//freopen("E:\\test.txt", "r", stdin);
	int n, m, k;
	while (~scanf("%d%d%d", &n, &k, &m)) {
		un.init(n);
		mp.clear();
		for (int i = 1; i <= n; i++) {
			scanf("%s", str);
			string s = str;
			mp[s] = i;
		}
		for (int i = 1; i <= n; i++)
			scanf("%d", &val[i]);
		int x, y, st;
		while (k--) {
			scanf("%d%d", &x, &st);
			for (int i = 1; i < x; i++) {
				scanf("%d", &y);
				un.unite(st, y);
			}
		}
		memset(gval, 0x3f, sizeof(gval));
		for (int i = 1; i <= n; i++) {
			gval[un.find(i)] = min(gval[un.find(i)], val[i]);
		}
		LL ans = 0;
		for (int i = 0; i < m; i++) {
			scanf("%s", str);
			string s = str;
			x = mp[s];
			y = un.find(x);
			ans += gval[y];
		}
		printf("%I64d\n", ans);
	}
	return 0;
}

你可能感兴趣的:(acm)