BZOJ 1191 超级英雄(二分图匹配)

把题目作为s集，锦囊作为t集。把每个题目和它可以用的锦囊连边，这样就构成了一个二分图，求出这个二分图最大匹配。

但是这个最大匹配有限制条件，就是对于每个可能的匹配集，如果s集的i点有匹配，那么i-1点一定有匹配。

具体实现，只需要将匈牙利算法稍微改动一下，如果当前没有找到增广路的话，就break。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())==‘-‘) flag=1;
    else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘;
    while((ch=getchar())>=‘0‘&&ch<=‘9‘)  res=res*10+(ch-‘0‘);
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=1005;
//Code begin...

struct Edge{int p, next;}edge[N<<1];
int head[N], cnt=1, linker[N], uN;
bool used[N];

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}
bool dfs(int u){
    for (int i=head[u]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (used[v]) continue;
        used[v]=true;
        if (linker[v]==-1||dfs(linker[v])){linker[v]=u; return true;}
    }
    return false;
}
int hungary(){
    int res=0;
    mem(linker,-1);
    FO(u,0,uN){
        mem(used,0);
        if (dfs(u)) res++;
        else break;
    }
    return res;
}
int main ()
{
    int n, m, u, v;
    scanf("%d%d",&n,&m);
    uN=m;
    FO(i,0,m) scanf("%d%d",&u,&v), add_edge(i,u), add_edge(i,v);
    printf("%d\n",hungary());
    return 0;
}