HDU 1151 Air Raid（最小路徑覆蓋 = 頂點數 - 最大匹配數）

Air Raid

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 Asia 2002, Dhaka (Bengal)
Recommend

題意：在一個城市，所有的街道連接方式都是單向的，但是每個街道都是從一個十字路口連接到另一個十字路口，已知說不存在回路。

一些傘兵要走遍小鎮的所有十字路口，且每個十字路口只有一個傘兵去。

求最少的傘兵數，即 最小路徑覆蓋

#include 
#include 
#include 
#include 
#include 
#include 
#define init(a) memset(a,0,sizeof(a))
#define PI acos(-1,0)
using namespace std;
const int maxn = 310;
const int maxm = 100001;
#define lson left, m, id<<1
#define rson m+1, right, id<<1|1
#define min(a,b) (a>b)?b:a
#define max(a,b) (a>b)?a:b
#define maxx 0x3f3f3f3f
const int N = 50010;
int ma[maxn][maxn];
int line[maxn];
bool vis[maxn];
int k,n,m;

int DFS(int u)
{
    for(int v = 1;v<=n;v++)
    {
        if(vis[v]==0 && ma[u][v]==1)
        {
            vis[v] = 1;
            if(line[v]==-1 || DFS(line[v]))
            {
                line[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
int K_M()
{
    int ans = 0;
    memset(line,-1,sizeof(line));
    for(int i = 1;i<=n;i++)
    {
        init(vis);
        ans+=DFS(i);
    }
    return ans;
}
int main()
{
    int x,y,z;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        init(ma);
        for(int i = 1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            ma[x][y] = 1;
        }
        int ans = K_M();
        //最小路徑覆蓋 = 頂點數 - 最大匹配數
        printf("%d\n",n-ans);
    }
    return 0;
}