1149 Dangerous Goods Packaging （25 分）

容器装货冲突检测

最新推荐文章于 2020-05-03 11:25:21 发布

原创最新推荐文章于 2020-05-03 11:25:21 发布 · 411 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲专栏收录该内容

53 篇文章

订阅专栏

本文介绍了一种算法，用于检测在集装箱装货时是否有可能发生因货物不兼容而导致的严重后果。通过输入一系列不可混装的货物对以及待检测的货物清单，算法能够判断清单内的货物是否全部兼容，避免潜在的危险。

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

本题需要注意的点，就是避免超时

题意：给你n组数据，每组数据代表他们在一起的话，会发生严重的后果，然后最下面是m组测试数组，每组数据给一个k值，代表本组测试数组有k个值，如果这组数据中出现n组数据中的任意一对，输出no，否则输出yes。把上题抽象变为数学题，其实题目说的是爆炸物，数据代表物品编号。

 #include<iostream>
 #include<string>
 #include<vector>
 #include<map>
 using namespace std;
 int main(){
 	int n,m;
 	map<int,vector<int>> mp;
 	cin >> n >> m;
 	for(int i= 0;i < n;i++){
 		int s1,s2;
 		scanf("%d %d",&s1,&s2);
 		mp[s1].push_back(s2);
 		mp[s2].push_back(s1);
	}
	while(m--){
		int k;
		cin >> k;
		int c[1000]={0};
		map<string,int> a;
		for(int i = 0;i < k;i++){
			scanf("%d",&c[i]);
			a[to_string(c[i])] = 1;
		}
		int flag = 0;
		for(int i = 0;i < k;i++){
				for(int j = 0;j < mp[c[i]].size();j++){
					if(a[to_string(mp[c[i]][j])] == 1){
						flag = 1;
						break;
					}
				}
			if(flag == 1){
				break;
			}
		}
		if(flag == 1){
			printf("No\n");
		}else{
			printf("Yes\n");
		}
	}
 	return 0;
 }