PAT 甲级 1149 Dangerous Goods Packaging

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#PAT甲级 #1149 #Dangerous Goods Packaging

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本文探讨了在使用容器运输货物时避免将不兼容的危险品混装的问题，通过算法检查每批货物列表，确保没有互斥的危险品被一同装载，以防潜在的安全隐患。

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1149 Dangerous Goods Packaging （25 point(s)）

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 (≤104), 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

经验总结：

emmmm 大概就是 PAT 甲级 1121 Damn Single 的加强版，只不过一个人可以和多个人配对，这里说人比较不法治，所以就改成了危险物品（手动滑稽），思路是一样的，存储所有的对，然后对于给出的所有物品，将不能与其共存的物品的flag置为true，然后从头至尾访问所有物品的flag，如果flag为true，说明危险物共存，如果遍历结束仍未发现，则安全~

AC代码

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int maxn=100010;
int n,m,a,b,k,goods[1010];
bool flag[maxn];
vector<int> couple[maxn];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;++i)
	{
		scanf("%d%d",&a,&b);
		couple[a].push_back(b);
		couple[b].push_back(a);
	}
	for(int i=0;i<m;++i)
	{
		scanf("%d",&k);
		memset(flag,false,sizeof(flag));
		for(int j=0;j<k;++j)
		{
			scanf("%d",&goods[j]);
			for(int x=0;x<couple[goods[j]].size();++x)
				flag[couple[goods[j]][x]]=true;
		}
		int f=true;
		for(int j=0;j<k;++j)
		{
			if(flag[goods[j]]==true)
			{
				f=false;
				break;
			}
		}	
		printf("%s\n",f?"Yes":"No");
	}
	return 0;
}