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 #include <stdio.h> #include <algorithm> #include <iostream> #include <map> #include <vector> #include <set> using namespace std; int n, m, k; set<int> adj[100001]; int inc[100001] = { 0 }; int req[110]; int main() {scanf("%d %d", &n, &m);for (int i = 0; i < n; i++) {int c1, c2;scanf("%d %d", &c1, &c2);adj[c1].insert(c2);adj[c2].insert(c1);inc[c1] = 1;inc[c2] = 1;}for (int i = 0; i < m; i++) {int flag = 0;scanf("%d", &k);for (int j = 0; j < k; j++) {scanf("%d", &req[j]);}for (int j = 0; j < k; j++) {if (flag == 0) {for (int q = j + 1; q < k; q++) {if (adj[req[j]].find(req[q]) != adj[req[j]].end()) {flag = 1;break;}}}else break;}printf("%s\n", flag == 0 ? "Yes" : "No");} }

注意點：看到題目第一眼想到的是圖的聯通塊，但想想第二題應該不會考圖，還是用map來做，用map好像有點麻煩，一個對應多個的時候不方便，那用vector，不好判斷在不在里面。最終還是用set比較好，就開一個很大的set數組，然后一個個去查找就好了，內存和時間都不超。

轉載于:https://www.cnblogs.com/tccbj/p/10412106.html

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