PAT Advanced—1149 Dangerous Goods Packaging (25分)

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

题目大意：
  给出一些不相容的危险品，然后进行装箱，看是否有不相容的危险品放在一起了

思路：
  保存用map，vector组合保存各种危险品对应不相容物品，然后对个集装箱的内容进行排序，对每一个物品对应的危险品取出，用二分法在集装箱中查找判断，找到相同为危险品输出No，否则Yes。

注意：

1. 二分查找binary_search()

代码：(C++)

#include
#include
#include
#include

using namespace std;

map<int,vector<int> > f;

int main()
{
	int n,m;
	cin>>n>>m;
	
	for(int i=0; i<n; i++)
	{
		int x,y;
		cin>>x>>y;
		f[x].push_back(y);
		f[y].push_back(x);
	}
	for(int i=0; i<m; i++)
	{
		int k;
		cin>>k;
		vector<int> v;
		for(int j=0; j<k; j++)
		{
			int x;
			cin>>x;
			v.push_back(x);
		}
		sort(v.begin(), v.end());
		int flag = 0;
		for(int j=0; j<k; j++)
		{
			for(int l=0; l<f[v[j]].size(); l++)
			{
				if(binary_search(v.begin(), v.end(), f[v[j]][l]) == true)
				{
					flag = 1;
					cout<<"No\n";
					break;
				}
			}
			if(flag == 1)
				break;
		}
		if(flag == 0)
			cout<<"Yes\n";
	}
	
	return 0;
}