(PAT 1149) Dangerous Goods Packaging (哈希)

港控/mmm° 2022-03-19 02:40 149阅读 0赞

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

解题思路:

该题类似于PAT 1121,利用哈希思想就可以解决

每件物品可能有多件不可相容的物品,我们利用vector把每件物品的不相容物品存入。

判断时,遍历每件物品的危险物品集,如果某件危险物品也在列表中,那么这列数据就输出为NO

#include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    const int MAXN = 100010;
    int N, M;
    vector<int> incompatible[MAXN];
    bool inbag[MAXN] = { false };
    
    int main() {
    	scanf("%d %d", &N, &M);
    	int pack1, pack2;
    	for (int i = 0; i < N; ++i) {
    		scanf("%d %d", &pack1, &pack2);
    		incompatible[pack1].push_back(pack2);
    		incompatible[pack2].push_back(pack1);
    	}
    	int num,tpack;
    	for (int i = 0; i < M; ++i) {
    		bool flag = true;
    		vector<int> packs;
    		scanf("%d", &num);
    		fill(inbag, inbag + MAXN, false);
    		for (int j = 0; j < num; ++j) {
    			scanf("%d", &tpack);
    			inbag[tpack] = true;
    			packs.push_back(tpack);
    		}
    		for (auto x : packs) {
    			for (auto y : incompatible[x]) {
    				if (inbag[y] == true) {
    					flag = false;
    					break;
    				}
    			}
    			if(!flag) break;
    		}
    		if (flag) {
    			printf("Yes\n");
    		}
    		else {
    			printf("No\n");
    		}
    	}
    	system("PAUSE");
    	return 0;
    }