zoukankan      html  css  js  c++  java
  • pat 1149 Dangerous Goods Packaging(25 分)

    1149 Dangerous Goods Packaging(25 分)

    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
     1 #include <map>
     2 #include <set>
     3 #include <queue>
     4 #include <cmath>
     5 #include <stack>
     6 #include <vector>
     7 #include <string>
     8 #include <cstdio>
     9 #include <cstring>
    10 #include <climits>
    11 #include <iostream>
    12 #include <algorithm>
    13 #define wzf ((1 + sqrt(5.0)) / 2.0)
    14 #define INF 0x3f3f3f3f
    15 #define LL long long
    16 using namespace std;
    17 const int MAX = 1e6 + 10;
    18 
    19 vector <int> ve[MAX];
    20 int book[MAX];
    21 
    22 int main()
    23 {
    24     freopen("Date.txt", "r", stdin);
    25     int n, m, a, b, k;
    26     scanf("%d%d", &n, &m);
    27     while (n --)
    28     {
    29         scanf("%d%d", &a, &b);
    30         ve[a].push_back(b);
    31         ve[b].push_back(a);
    32     }
    33     while (m --)
    34     {
    35         scanf("%d", &k);
    36         bool flag = false;
    37         memset(book, 0, sizeof(book));
    38         while (k --)
    39         {
    40             scanf("%d", &a);
    41             if (flag) continue;
    42             if (book[a])
    43             {
    44                 flag = true;
    45                 printf("No
    ");
    46                 continue;
    47             }
    48             for (int i = 0; i < ve[a].size(); ++ i)
    49                 book[ve[a][i]] = 1;
    50         }
    51         if (!flag) printf("Yes
    ");
    52     }
    53     return 0;
    54 }
  • 相关阅读:
    路由基础、多app共存,路由分配、路由分发(将app自己的路由分发给应用自身管理)、反解
    Django项目的创建与介绍,三件套,静态文件,配置Mysql完成数据迁移,单表ORM记录的增删改查
    Django框架导读
    Flask简易版本、Ajax、DOM操作,表单操作
    JQuery
    0820-信心赛
    codeforces比赛总(吐)结(嘈)
    洛谷P3403 跳楼机(最短路)
    求逆序对的三种方法
    NKOJ 3751 扫雷游戏
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9612468.html
Copyright © 2011-2022 走看看