1149 Dangerous Goods Packaging（25 point(s)）

最新推荐文章于 2020-07-23 08:55:21 发布

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最新推荐文章于 2020-07-23 08:55:21 发布

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分类专栏： PAT甲级算法与数据结构 STL 文章标签： PAT甲级 STL

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本文介绍了一个算法问题，涉及判断一组商品中是否包含不兼容的商品，特别是当这些商品需要装入同一容器进行运输时。通过输入成对的不兼容商品列表及待检查的商品组合，算法能够确定是否存在冲突。

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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 (≤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

//AC code
/*
 1:采用 map<int,vector<int> > 很容易解决 
*/ 
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main(){
    int n,k,c1,c2;
    cin>>n>>k;
    map<int,vector<int> > m; //很有趣,vector<int>和>间要加空格 ,否则会收到一个错误 
    for (int i=0;i<n;i++) {
        cin>>c1>>c2;
        m[c1].push_back(c2);
        m[c2].push_back(c1);
    }
    for(int i=0;i<k;i++){
        int cnt, tag = 0;
        bool a[100000] = {false};
        cin>>cnt;
        vector<int>v(cnt); 
        for (int i = 0; i < cnt; i++) {
            cin>>v[i];
            a[v[i]] = true;
        }
        for (int i=0;i<cnt;i++)
            for (int j=0;j<m[v[i]].size();j++)
                if (a[m[v[i]][j]] == true) tag = 1;
        cout<<(tag ?"No" : "Yes")<<endl;
    }
    return 0;
}