zoukankan      html  css  js  c++  java
  • UVA10305 Ordering Tasks

    题目链接:https://cn.vjudge.net/problem/UVA-10305(忍不住uva连接满)

    John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.

    Input

    The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and mn is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.

    Output

    For each instance, print a line with n integers representing the tasks in a possible order of execution.

    Sample Input

    5 4
    1 2
    2 3
    1 3
    1 5
    0 0

    Sample Output

    1 4 2 5 3

     

    ***************************************************************************************

    题意:给你n个点(从1开始),还有要求一些点i必须在j前面。输出任意一种排序。

    题解:裸的拓扑排序,直接套模板。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <string>
     6 #include <vector>
     7 #include <map>
     8 #include <set>
     9 #include <queue>
    10 #include <sstream>
    11 #include <algorithm>
    12 using namespace std;
    13 #define pb push_back
    14 #define mp make_pair
    15 #define ms(a, b)  memset((a), (b), sizeof(a))
    16 //#define LOCAL
    17 typedef long long LL;
    18 const int inf = 0x3f3f3f3f;
    19 const int maxn = 100+10;
    20 const int mod = 1e9+7;
    21 int gap[maxn][maxn];
    22 int topo[maxn], c[maxn], t;
    23 int n, m;
    24 bool dfs(int u){
    25     c[u] = -1;
    26     for(int v = 1; v<=n; v++)   if(gap[u][v]){
    27         if(c[v]< 0) return false;
    28         else if(!c[v] && !dfs(v) )  return false;
    29     }
    30     c[u] = 1;
    31     topo[--t] = u;
    32     return true;
    33 }
    34 bool toposort(){
    35     t = n+1;
    36     ms(c, 0);
    37     for(int u = 1; u<=n ;u++)   if(!c[u])
    38         if(!dfs(u)) return false;
    39     return true;
    40 }
    41 int main()
    42 {
    43     #ifdef LOCAL
    44         freopen("input.txt" , "r", stdin);
    45     #endif // LOCAL
    46     while(~scanf("%d%d", &n, &m)){
    47         if(n==0 && m==0)    break;
    48         ms(gap, 0);
    49         for(int i = 0;i<m;i++){
    50             int a, b;
    51             scanf("%d%d", &a, &b);
    52             gap[a][b] = 1;
    53         }
    54         if(toposort()){
    55             for(int i =1 ; i<=n ;i++)
    56                 if(i==1)    printf("%d", topo[i]);
    57                 else    printf(" %d", topo[i]);
    58             printf("
    ");
    59         }
    60     }
    61     return 0;
    62 }
    View Code


  • 相关阅读:
    [原创]浅谈测试团队转型,思维模式的转变是关键
    [原创]浅谈IT人如何做理财规划
    [原创]浅谈对华为34岁以上员工“退休”
    [原创]2017年1月读书推荐
    [原创] 上海利得基金招聘测试经理/测试工程师/测试开发工程师(长期有效)
    [原创]浅谈从“述职”向“述能”转变
    IBM的IT战略规划方法论
    腾讯企业大学培训经验
    [原创]Android App上线检查checklist
    移动开发规范
  • 原文地址:https://www.cnblogs.com/denghaiquan/p/6668417.html
Copyright © 2011-2022 走看看