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  • POJ 1325 Machine Schedule【最小点覆盖】

    E - Machine Schedule
    Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    Description

    As we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. Scheduling problems differ widely in the nature of the constraints that must be satisfied and the type of schedule desired. Here we consider a 2-machine scheduling problem. 

    There are two machines A and B. Machine A has n kinds of working modes, which is called mode_0, mode_1, ..., mode_n-1, likewise machine B has m kinds of working modes, mode_0, mode_1, ... , mode_m-1. At the beginning they are both work at mode_0. 

    For k jobs given, each of them can be processed in either one of the two machines in particular mode. For example, job 0 can either be processed in machine A at mode_3 or in machine B at mode_4, job 1 can either be processed in machine A at mode_2 or in machine B at mode_4, and so on. Thus, for job i, the constraint can be represent as a triple (i, x, y), which means it can be processed either in machine A at mode_x, or in machine B at mode_y. 

    Obviously, to accomplish all the jobs, we need to change the machine's working mode from time to time, but unfortunately, the machine's working mode can only be changed by restarting it manually. By changing the sequence of the jobs and assigning each job to a suitable machine, please write a program to minimize the times of restarting machines. 

    Input

    The input file for this program consists of several configurations. The first line of one configuration contains three positive integers: n, m (n, m < 100) and k (k < 1000). The following k lines give the constrains of the k jobs, each line is a triple: i, x, y. 

    The input will be terminated by a line containing a single zero. 

    Output

    The output should be one integer per line, which means the minimal times of restarting machine.

    Sample Input

    5 5 10
    0 1 1
    1 1 2
    2 1 3
    3 1 4
    4 2 1
    5 2 2
    6 2 3
    7 2 4
    8 3 3
    9 4 3
    0
    

    Sample Output

    3
    之所以把题目粘下来是因为自己读题不仔细wa一次
    大意:
    有两台机器分别有n,m种模式,开始的时候全都在0模式
    告诉你有一些任务可以在第一台机器的x1模式下完成也可以在第二台机器的x2模式下完成
    并且每台机器切换模式需要重启
    问最少重启多少次

    分析:
    左右集合分别为两台机器的模式
    边为task
    那么只要选取最少的点覆盖所有的边即可
    注意一点是一开始在0模式
    所以只要对于0不建边即可
    网络流亦可以做

    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    using namespace std;
    
    const int maxn = 105;
    int n;
    
    vector<int> G[maxn];
    int vis[maxn];
    int Link[maxn];
    bool Find(int u) {
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i];
            if(!vis[v]) {
                vis[v] = 1;
                if(Link[v] == -1 || Find(Link[v])) {
                    Link[v] = u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int solve() {
        int cnt = 0;
        memset(Link, -1, sizeof(Link));
        for(int i = 0; i < n; i++) {
            memset(vis, 0, sizeof(vis));
            if(Find(i)) cnt++;
        }
        return cnt;
    }
    
    int main() {
        int m, k;
        int a, b, c;
        while(scanf("%d",&n) && n) {
            scanf("%d %d",&m, &k);
            n = max(n, m);
            for(int i = 0; i < n; i ++) G[i].clear();
            for(int i = 0; i < k; i++) {
                scanf("%d %d %d",&a, &b, &c);
                if(b == 0 || c == 0) continue;
                G[b].push_back(c);
            }
            printf("%d
    ",solve());
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/zhanzhao/p/3920310.html
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