PAT_A1149#Dangerous Goods Packaging-白红宇

PAT_A1149#Dangerous Goods Packaging

阅读量：6585 次

发布时间：2019-06-24

本文共 2204 字，大约阅读时间需要 7 分钟。

Source:

Description:

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:

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 ( $\leq) is the number of goods and G[i]'s are the IDs of the goods. To make it simple, each good is represented by a 5-digit ID number. All the numbers in a line are separated by spaces.$

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 320001 2000220003 2000420005 2000620003 2000120005 2000420004 200064 00001 20004 00002 200035 98823 20002 20003 20006 100103 12345 67890 23333

Sample Output:

NoYesYes

Keys:

散列（Hash）

Attention:

若1个元素与N个元素互斥，属于散列；若一系列元素相互排斥的话，则是并查集

Code:

1 /* 2 Data: 2019-05-15 20:14:44 3 Problem: PAT_A1149#Dangerous Goods Packaging 4 AC: 32:04 5  6 题目大意： 7 给出互斥物体的列表，和需要运输的货物，判断这些货物是否可以共存 8 输入： 9 第一行，互斥物体总数N<=1e4，运输货物总数M<=100；10 接下来N行，给出互斥对（一个物品可能有多个互斥物品），5位数字表示；11 接下来M行，首先给出容量K<=1e3，和K个物品12 */13 14 #include
     
      15 #include
      
       16 #include
       
        17 using namespace std;18 const int M=110,N=1e6;19 int mp[N],goods[M];20 vector
        
          pairs[N];21 22 int main()23 {24 #ifdef    ONLINE_JUDGE25 #else26     freopen("Test.txt", "r", stdin);27 #endif28 29     int n,m,v1,v2,k;30     scanf("%d%d", &n, &m);31     for(int i=0; i