• hdu 5154 Harry and Magical Computer


    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=5154

    Harry and Magical Computer

    Description

    In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

    Input

    There are several test cases, you should process to the end of file.
    For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. $1 leq n leq 100,1 leq m leq 10000$
    The next following m lines, each line contains two numbers a b, indicates a dependencies $(a, b). 1 leq a, b leq n$

    Output

    Output one line for each test case. 
    If the computer can finish all the process print "YES" (Without quotes).
    Else print "NO" (Without quotes).

    Sample Input

    3 2
    3 1
    2 1
    3 3
    3 2
    2 1
    1 3

    Sample Output

    YES
    NO

    先建立一张有向图,再遍历,若有顶点未访问到输出"NO",否则输出"YES"。。。

     1 #include<algorithm>
     2 #include<iostream>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<vector>
     7 #include<queue>
     8 #include<map>
     9 using std::cin;
    10 using std::cout;
    11 using std::endl;
    12 using std::find;
    13 using std::sort;
    14 using std::pair;
    15 using std::vector;
    16 using std::queue;
    17 using std::multimap;
    18 #define pb(e) push_back(e)
    19 #define sz(c) (int)(c).size()
    20 #define mp(a, b) make_pair(a, b)
    21 #define all(c) (c).begin(), (c).end()
    22 #define iter(c) decltype((c).begin())
    23 #define cls(arr,val) memset(arr,val,sizeof(arr))
    24 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    25 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
    26 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
    27 const int N = 110;
    28 int n, m, G[N][N];
    29 bool vis[N], flag[N];
    30 typedef unsigned long long ull;
    31 void bfs() {
    32     bool f = false;
    33     queue<int> que;
    34     rep(i, n) {
    35         if (!flag[i]) {
    36             vis[i] = true;
    37             que.push(i);
    38         }
    39     }
    40     while (!que.empty()) {
    41         int p = que.front(); que.pop();
    42         rep(i, n) {
    43             if (G[p][i]) {
    44                 if (vis[i]) continue;
    45                 que.push(i);
    46                 vis[i] = true;
    47             }
    48         }
    49     }
    50     rep(i, n) {
    51         if (!vis[i]) { f = true; break; }
    52     }
    53     puts(f ? "NO" : "YES");
    54 }
    55 int main() {
    56 #ifdef LOCAL
    57     freopen("in.txt", "r", stdin);
    58     freopen("out.txt", "w+", stdout);
    59 #endif
    60     int a, b;
    61     while (~scanf("%d %d", &n, &m)) {
    62         cls(vis, 0), cls(flag, 0), cls(G, 0);
    63         rep(i, m) {
    64             scanf("%d %d", &a, &b);
    65             G[b - 1][a - 1] = 1;
    66             flag[a - 1] = 1;
    67         }
    68         bfs();
    69     }
    70     return 0;
    71 }
    View Code
    By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明
  • 相关阅读:
    JavaScript变量和作用域
    遥感专业词汇
    linux修改文件所属用户和用户组
    当singleton Bean依赖propotype Bean,可以使用在配置Bean添加look-method来解决
    linux中的目录和文件的统计
    linux命令在文件中根据命令查找
    走进ELK原理
    nohub和重定向文件
    HashMap与TreeMap按照key和value排序
    List自定义排序
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4603743.html
Copyright © 2020-2023  润新知