HDU1511 Air Raid(二分图，最小路径覆盖)

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

2433 41 32 3331 31 22 3

Sample Output

21

思路

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

最小路径覆盖问题吗，只需要让顶点数减去最大匹配数即可

代码

#include<cstdio>#include<cstring>#include<string>#include<set>#include<iostream>#include<stack>#include<queue>#include<vector>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define ll long longusing namespace std;const int N=1600+20;int vis[N],match[N],n,m;vector<int>G[N];int dfs(int u){    for(int i=0; i<G[u].size(); i++)    {        int v=G[u][i];        if(!vis[v])        {            vis[v]=1;            if(!match[v]||dfs(match[v]))            {                match[v]=u;                return 1;            }        }    }    return 0;}int query(){    mem(match,0);    int sum=0;    for(int i=1; i<=n; i++)    {        mem(vis,0);        if(dfs(i))sum++;    }    return sum;}int main(){    int t,a,b;    scanf("%d",&t);    while(t--)    {        mem(G,0);        scanf("%d%d",&n,&m);        while(m--)        {            scanf("%d%d",&a,&b);            G[a].push_back(b);        }        printf("%d\n",n-query());    }    return 0;}