1149 Dangerous Goods Packaging

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本文探讨了在货物运输中避免将不兼容的危险品放置在同一容器内的必要性，特别是氧化剂与易燃液体混装可能导致爆炸的问题。通过输入危险品配对和待运货物清单，算法判断是否所有货物可以安全地装在同一容器内。

1149 Dangerous Goods Packaging
When shipping goods with containers, we have to be careful not to pack some incompatible goods into the same container, or we might get ourselves in serious trouble. For example, oxidizing agent （氧化剂） must not be packed with flammable liquid （易燃液体）, or it can cause explosion.

Now you are given a long list of incompatible goods, and several lists of goods to be shipped. You are supposed to tell if all the goods in a list can be packed into the same container.

Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: N (≤10
4
), the number of pairs of incompatible goods, and M (≤100), the number of lists of goods to be shipped.

Then two blocks follow. The first block contains N pairs of incompatible goods, each pair occupies a line; and the second one contains M lists of goods to be shipped, each list occupies a line in the following format:

K G[1] G[2] … G[K]
where K (≤1,000) is the number of goods and G[i]'s are the IDs of the goods. To make it simple, each good is represented by a 5-digit ID number. All the numbers in a line are separated by spaces.

Output Specification:
For each shipping list, print in a line Yes if there are no incompatible goods in the list, or No if not.

Sample Input:
6 3
20001 20002
20003 20004
20005 20006
20003 20001
20005 20004
20004 20006
4 00001 20004 00002 20003
5 98823 20002 20003 20006 10010
3 12345 67890 23333
Sample Output:
No
Yes
Yes
分析：有些物品无法共存，放在一起就会导致危险情况的出现。
先给出一些不能物品的药品对，给出放在一起的物品的编号，判断这些物品能否放在一起。
参考代码：

#include<iostream>
#include<vector>
#include<map>
#include<cstring>
#define N 100006
using namespace std;
int t[N];
int main()
{
	int n, m, a, b, k, num, flag;
	vector<vector<int>>v(N);
	scanf_s("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf_s("%d%d", &a, &b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for (int i = 1; i <= m; i++) {
		scanf_s("%d", &k);
		flag = 1;
		memset(t, -1, sizeof(t));
		for (int j = 1; j <= k; j++) {
			scanf_s("%d", &a);
			if (flag) {
				for (int p = 0; p < v[a].size(); p++) {
					if (t[v[a][p]] != -1)flag = 0;
				}
				t[a] = 1;
			}
		}
		if (flag)cout << "Yes" << endl;
		else cout << "No" << endl;
	}
    return 0;
}