HDOJ---1151 Air Raid[匈牙利算法：最小路径覆盖数=原图顶点数–二分图最大匹配数]

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1859    Accepted Submission(s): 1191

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 parat

 

rooper 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

Asia 2002, Dhaka (Bengal)

 

 

Recommend

Ignatius.L

 

 

 

 

 

 

 

 

 

 

 

 

哎，比赛后才做出来，还不够，多做题吧 。

code：

 1 #include <iostream>   

 2 #include <iomanip>   

 3 #include <fstream>   

 4 #include <sstream>   

 5 #include <algorithm>   

 6 #include <string>   

 7 #include <set>   

 8 #include <utility>   

 9 #include <queue>   

10 #include <stack>   

11 #include <list>   

12 #include <vector>   

13 #include <cstdio>   

14 #include <cstdlib>   

15 #include <cstring>   

16 #include <cmath>   

17 #include <ctime>   

18 #include <ctype.h> 

19 using namespace std;

20 

21 #define MAXN 130

22 

23 int map[MAXN][MAXN];

24 int path[MAXN];

25 int vst[MAXN];

26 int n;

27 

28 bool dfs(int v)

29 {

30     for(int i=1;i<=n;i++)

31         if(!vst[i]&&map[v][i])

32         {

33             vst[i]=1;

34             if(path[i]==-1||dfs(i))

35             {

36                 path[i]=v;

37                 return true;

38             }

39         }

40     return false;

41 }

42 

43 int hungary()

44 {

45     memset(path,-1,sizeof(path));

46     int cnt=0;

47     for(int i=1;i<=n;i++)

48     {

49         memset(vst,0,sizeof(vst));

50         if(dfs(i))

51             cnt++;

52     }

53     return cnt;

54 }

55 

56 int main()

57 {

58     int t;

59     int k;

60     int i;

61     int a,b;

62     scanf("%d",&t);

63     while(t--)

64     {

65         memset(map,0,sizeof(map));

66         scanf("%d%d",&n,&k);

67         for(i=1;i<=k;i++)

68         {

69             scanf("%d%d",&a,&b);

70             map[a][b]=1;

71         }

72         int ans=n-hungary();

73         printf("%d\n",ans);

74     }

75     return 0;

76 }

 

 

 

 

 

 

 

 

 

法四 拆点法
拆点法是用于解决最小路径覆盖问题的，给出一个图
要找到几条路径，可以把所有的点经过，并且路径之间不可以交叉。我们的做法是把点拆成两部分（点1 拆为x1,y1. 点2拆为x2,y2……）
如果我们对这个图求二分图的最大匹配，你会发现每个匹配对应着一个路径覆盖，因此，此二分图的最大匹配即：原图中的最小路径覆盖上的边的个数（路径是由0条，1条或多条边组成的）。那么原图的最小路径覆盖数 = 原图顶点数 – 最小路径上的边数 也就是 原图的最小路径覆盖数 = 原图顶点数 – 二分图最大匹配数。