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 （氧化劑） must not be packed with flammable liquid （易燃液體）, 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

題意：碼頭上在進行集裝箱裝運時，有些物品時不能放在一起運輸的，例如氧化劑和易燃液體不能放在一起運輸。現給定n對相斥的物品，然後給你m張運輸清單，問題這m張清單能否進行運輸，可以的話輸出Yes，否則輸出No。

思路：使用unordered_map<int,vector<int>>把所有相斥的物品放在一個對應的vector裡，然後對每張運輸清單逐一掃描， 若當前物品的相斥物品存在輸出No，否則輸出Yes。

參考程式碼：

#include<cstdio>
#include<vector>
#include<cstring>
#include<unordered_map>
using namespace std;
const int maxn=100000;
bool flag[maxn];
unordered_map<int,vector<int>> mp;
int n,m,u,v,t,k;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d%d",&u,&v);
        mp[u].push_back(v);
        mp[v].push_back(u);
    }
    for(int i=0;i<m;i++){
        scanf("%d",&k);
        bool tag=true;
        memset(flag,false,sizeof(flag));
		vector<int> v(k);
        for(int j=0;j<k&&tag;j++){
            scanf("%d",&v[j]);
			flag[v[j]]=true;
		}
		for(int j=0;j<k&tag;j++){
            for(int l=0;l<mp[v[j]].size()&&tag;l++){
                u=mp[v[j]][l];
                if(flag[u]) tag=false;
            }
		}
        printf("%s\n",tag?"Yes":"No");
    }
    return 0;
}