zoukankan      html  css  js  c++  java
  • HDU-1151

    Air Raid

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3661    Accepted Submission(s): 2408


    Problem Description
    Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's streets form no cycles.

    With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper.
     
    Input
    Your program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format:

    no_of_intersections
    no_of_streets
    S1 E1
    S2 E2
    ......
    Sno_of_streets Eno_of_streets

    The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town's streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) - the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) - the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections.

    There are no blank lines between consecutive sets of data. Input data are correct.
     
    Output
    The result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections in the town.
     
    Sample Input
    2
    4
    3
    3 4
    1 3
    2 3
    3
    3
    1 3
    1 2
    2 3
    Sample Output
    2
    1
     
    Source
     
    Recommend
    Ignatius.L
    /**
              题意:有向无环图的最小路径覆盖
              做法:二分图最大匹配 有向无环图的最小路径覆盖 = 该图的顶点数-该图的最大匹配。
    **/
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<string.h>
    #include<stdio.h>
    #define maxn 220
    using namespace std;
    int g[maxn][maxn];
    int linker[maxn];
    int used[maxn];
    int n,m;
    int dfs(int u)
    {
        for(int v = 0; v<n; v++)
        {
            if(g[u][v] && used[v] == 0)
            {
                used[v] = 1;
                if(linker[v] || dfs(linker[v]))
                {
                    linker[v] = u;
                    return 1;
                }
            }
        }
        return 0;
    }
    int hungary()
    {
        int res = 0;
        memset(linker,-1,sizeof(linker));
        for(int i=0; i<n; i++)
        {
            memset(used,0,sizeof(used));
            res += dfs(i);
        }
        return res;
    }
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
        int T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d %d",&n,&m);
            int u,v;
            memset(g,0,sizeof(g));
            for(int i=0; i<m; i++)
            {
                scanf("%d %d",&u,&v);
                u--;
                v--;
                g[u][v] = 1;
            }
            int res = hungary();
            printf("%d
    ",n - res);
        }
        return 0;
    }
  • 相关阅读:
    Effective C# 学习笔记(二十九)在范型中的协变和逆变
    Effective C# 学习笔记(二十一)为类型定义有限的职责
    Effective C# 学习笔记(十四) 尽量减少重复性的初始化逻辑
    Effective C# 学习笔记(十三)对静态类成员使用合适的初始化方式
    Effective C# 学习笔记(二十六)防止返回类内部的对象引用
    Effective C# 学习笔记(十二) 多用成员变量初始化,少用指定赋值
    Effective C# 学习笔记(二十七)使你的类型可被序列化
    Effective C# 学习笔记(二十四)运用Delegates来实现回调
    Effective C# 学习笔记(二十)不可变的原子值类型的好处
    UIPageController与UIScrollView的联合使用
  • 原文地址:https://www.cnblogs.com/chenyang920/p/4392745.html
Copyright © 2011-2022 走看看