poj 1422 Air Raid （最小路径覆盖白痴题）

Air Raid

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8141		Accepted: 4877

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

题目大意：有n个点和m条有向边，现在要在点上放一些伞兵，然后伞兵沿着图走，直到不能走为止
每条边只能是一个伞兵走过，问最少放多少个伞兵

思路：一看就是最小路径覆盖，裸题。。

将有向图无环图转为二分图，可以这么理解，每个点拆成2个点，一个只进，一个只出，左面1-n个点代表只出的点，右面1-n个点代表只进的点，从左面点连向右面的点， 从而达到了有向图；

关于更详细的解答：转自百度百科

PXP的有向图中，路径覆盖就是在图中找一些路径，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）；如果不考虑图中存在回路，那么每条路径就是一个弱连通子集．

由上面可以得出：

1.一个单独的顶点是一条路径；

2．如果存在一路径p1,p2,......pk，其中p1 为起点，pk为终点，那么在覆盖图中，顶点p1,p2,......pk不再与其它的顶点之间存在有向边．

对于一个路径覆盖，有如下性质：

1、每个顶点属于且只属于一个路径。

2、路径上除终点外，从每个顶点出发只有一条边指向路径上的另一顶点。

路径覆盖与 二分图匹配的关系（必须是有向无环图）：

最小路径覆盖=|P|－最大匹配数

其中最大匹配数的求法是把P中的每个顶点pi分成两个顶点pi'与pi''，如果在p中存在一条pi到pj的边，那么在二分图P'中就有一条连接pi'与pj''的无向边；这里pi' 就是p中pi的出边，pj''就是p中pj 的一条入边；

对于 公式：最小路径覆盖=|P|－最大匹配数；可以这么来理解；

如果匹配数为零，那么P中不存在有向边，于是显然有：

最小路径覆盖=|P|－最大匹配数=|P|－0=|P|；即P的最小路径覆盖数为|P|；

P'中不在于匹配边时，路径覆盖数为|P|；

如果在P'中增加一条匹配边pi'－－>pj''，那么在图P的路径覆盖中就存在一条由pi连接pj的边，也就是说pi与pj 在一条路径上，于是路径覆盖数就可以减少一个；

如此继续增加匹配边，每增加一条，路径覆盖数就减少一条；直到匹配边不能继续增加时，路径覆盖数也不能再减少了，此时就有了前面的公式；但是这里只 是说明了每条匹配边对应于路径覆盖中的一条路径上的一条连接两个点之间的有向边；下面来说明一个路径覆盖中的每条连接两个顶点之间的有向边对应于一条匹配 边；

与前面类似，对于路径覆盖中的每条连接两个顶点之间的每条有向边pi--->pj，我们可以在匹配图中对应做一条连接pi'与pj''的边， 显然这样做出来图的是一个匹配图（这一点用反证法很容易证明，如果得到的图不是一个匹配图，那么这个图中必定存在这样两条边 pi'---pj'' 及 pi' ----pk''，（j!=k），那么在路径覆盖图中就存在了两条边pi-->pj, pi--->pk ，那边从pi出发的路径就不止一条了，这与路径覆盖图是矛盾的；还有另外一种情况就是存在pi'---pj'',pk'---pj''，这种情况也类似可证）；

至此，就说明了匹配边与路径覆盖图中连接两顶点之间边的一一对应关系，那么也就说明了前面的公式成立！

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 505;
vector v[maxn];
int n, m, k, book[maxn], match[maxn];
int Find(int x)
{
    for(int i = 0; i < v[x].size(); i++)
    {
        int to = v[x][i];
        if(book[to]) continue;
        book[to] = 1;
        if(!match[to] || Find(match[to]))
        {
            match[to] = x;
            return 1;
        }
    }
    return 0;
}
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int x, y, ans = 0;
        memset(match, 0, sizeof(match));
        for(int i = 0; i < maxn; i++)
            v[i].clear();
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &x, &y);
            v[x].push_back(y);
        }
        for(int i = 1; i <= n; i++)
        {
            memset(book, 0, sizeof(book));
            ans += Find(i);
        }
        printf("%d\n", n-ans);
    }
    return 0;
}

Air Raid

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8141		Accepted: 4877

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