POJ 1422 Air Raid

Air Raid

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6275		Accepted: 3756

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

Dhaka 2002

最小路径覆盖问题

首先说下最小路径覆盖和最小边覆盖的区别吧：

　　最小边覆盖：设图G=(V,E)，求一个元素个数最少的边集合F，使G中任意顶点都至少是F中某条边的端点。做法是利用：最大匹配+最小边覆盖=顶点数

　　最小路径覆盖：在一个有向图中找一些路径，使之覆盖了图中的所有顶点，且任何一个顶点有且只有一条路径与之关联（如果把这些路径中的每条路径从它的起始点走到它的终点，那么恰好可以经过图中的每个顶点一次且仅一次）。有向无环图的中的做法是：根据原图构造二分图，构造方法是将每个顶点一分为二，即将顶点i分为i1和i2，若原图中存在一条从i到j的有向边，则在新二分图中从i1到j2边接一条边；接下来利用：二分图中最大匹配+原图最小路径覆盖=原图顶点数

这道题中只要先根据所给图构造二分图，然后在二分图中套用匈牙利算法求出最大匹配，再利用上式求解即可

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#define MAX_V 250

using namespace std;

int V,E;
int match[MAX_V];       //所匹配的顶点
bool used[MAX_V];       //DFS中用到 的访问标记
vector<int> G[MAX_V];   //图的邻接表表示

//向图中增加一条边接u和v的边
void add_edge(int u,int v)
{
    G[u].push_back(v);
    G[v].push_back(u);
}

//通过DFS寻找增广路
bool dfs(int v)
{
    used[v]=true;

    for(int i=0;i<G[v].size();i++)
    {
        int u=G[v][i],w=match[u];
        if(w<0||(!used[w]&&dfs(w)))
        {
            match[v]=u;
            match[u]=v;
            return true;
        }
    }

    return false;
}

//求解二分图的最大匹配
int bipartite_matching()
{
    int res=0;

    memset(match,-1,sizeof(match));

    for(int v=1;v<=V;v++)
    {
        if(match[v]<0)
        {
            memset(used,false,sizeof(used));
            if(dfs(v))
                res++;
        }
    }

    return res;
}

int main()
{
    int kase;

    scanf("%d",&kase);

    while(kase--)
    {
        for(int i=1;i<=2*V;i++)
            G[i].clear();

        scanf("%d %d",&V,&E);

        int a,b;
        for(int i=0;i<E;i++)
        {
            scanf("%d %d",&a,&b);
            add_edge(a,V+b);
        }

        printf("%d
",V-bipartite_matching());
    }

    return 0;
}