POJ-1325 Machine Schedule 二分图匹配 最小点覆盖问题

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------showtime----------------------*/
            const int maxn = 1e5+9;
            struct edge{
                int v,nx;
            }e[maxn];
            int h[maxn],tot = 0;
            void addedge(int u,int v){
                e[tot].v = v;
                e[tot].nx = h[u];
                h[u] = tot++;
            }
            int mx[maxn],my[maxn],vis[maxn];
            bool dfs(int x){

                for(int i = h[x]; ~i; i = e[i].nx){
                    int v = e[i].v;

                    if(vis[v]==0){
                            vis[v] = 1;
                            if(mx[v]==-1||dfs(mx[v])){
                                    mx[v] = x;
                                    my[x] = v;
                                    return true;
                            }
                    }
                }
                return false;
            }
int main(){
            int n,m,k;
            while(~scanf("%d", &n) && n){
                tot = 0;
                memset(h,-1,sizeof(h));
                memset(mx,-1,sizeof(mx));
                memset(my,-1,sizeof(my));

                scanf("%d%d", &m, &k);
                for(int i=1; i<=k; i++){
                    int u,v,q;
                    scanf("%d%d%d", &q, &u, &v);
                    if(u*v)addedge(u,v);
                }

                int ans = 0;
                for(int i=1; i<n; i++){
                    memset(vis,0,sizeof(vis));    
                    if(dfs(i))ans++;
                }
                printf("%d\n", ans);
            }            
            return 0;
}