POJ 1422：Air Raid（最大獨立集）

Air Raid

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 6547 Accepted: 3896

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

最小路徑覆蓋：

用盡可能少的不相交簡單路徑覆蓋有向無環圖Ｇ的全部結點。解決此類問題能夠創建一個二分圖模型。把全部頂點i拆成兩個：Ｘ結點集中的i和Y結點集中的i',若是有邊i->j，則在二分圖中引入邊i->j'，設二分圖最大匹配爲m,則結果就是n-m。

01.#include<cstdio>  
02.#include<cstring>  
03.#include<algorithm>  
04.#include<iostream>  
05.  
06.using namespace std;  
07.  
08.const int M = 1000 + 5;  
09.int t, n, k;  
10.int link[M];  
11.bool MAP[M][M];  
12.int cover[M];  
13.int ans;  
14.  
15.void init()  
16.{  
17.    int x, y;  
18.    memset(MAP, false, sizeof(MAP));  
19.    for(int i=1; i<=k; i++)  
20.    {  
21.        scanf("%d%d", &x, &y);  
22.        MAP[x][y]=true;  
23.    }  
24.}  
25.  
26.bool dfs(int x)  
27.{  
28.    for(int y=1; y<=n; y++)  
29.    {  
30.        if(MAP[x][y] && !cover[y])  
31.        {  
32.            cover[y]=true;  
33.            if(!link[y] || dfs(link[y]))  
34.            {  
35.                link[y]=x;  
36.                return true;  
37.            }  
38.        }  
39.    }  
40.    return false;  
41.}  
42.  
43.int main ()  
44.{  
45.    scanf("%d", &t);  
46.    while( t-- )  
47.    {  
48.        scanf("%d%d", &n, &k);  
49.        ans=0;  
50.        init();  
51.        memset(link, 0, sizeof(link));  
52.        for(int i=1; i<=n; i++)  
53.        {  
54.            memset(cover, false, sizeof(cover));  
55.            if( dfs(i) )  
56.                ans++;  
57.        }  
58.        printf("%d\n", n-ans);  
59.    }  
60.  
61.    return 0;  
62.}