bzoj2718 [Violet 4]毕业旅行

Description

Input

Output

最多可选多少景点

Sample Input

7 6
1 2
2 3
5 4
4 3
3 6
6 7

Sample Output

2

HINT

 

 

这题是结论题

答案=最长反链=最小路径覆盖=n-二分图最大匹配

先floyd处理出两点之间的联通性，然后拆点，如果A能到B则A向B'连边

#include<cstdio>
#include<iostream>
#include<cstring>
#define LL long long
#define inf 0x7ffffff
#define S 0
#define T 2*n+1
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct edge{int to,next,v;}e[100010];
int n,m,cnt=1,ans;
bool go[1010][1010];
int head[100010];
int q[100010];
int h[100010];
inline void ins(int u,int v,int w)
{
    e[++cnt].to=v;
    e[cnt].v=w;
    e[cnt].next=head[u];
    head[u]=cnt;
}
inline void insert(int u,int v,int w)
{
    ins(u,v,w);
    ins(v,u,0);
}
inline bool bfs()
{
    memset(h,-1,sizeof(h));
    q[1]=S;h[S]=0;int t=0,w=1;
    while (t<w)
    {
        int now=q[++t];
        for (int i=head[now];i;i=e[i].next)
          if (e[i].v&&h[e[i].to]==-1)
          {
            q[++w]=e[i].to;
            h[e[i].to]=h[now]+1;
          }
    }
    if (h[T]==-1)return 0;
    return 1;
}
inline int dfs(int x,int f)
{
    if (x==T||!f)return f;
    int w,used=0;
    for (int i=head[x];i;i=e[i].next)
        if (e[i].v&&h[e[i].to]==h[x]+1)
        {
            w=dfs(e[i].to,min(f-used,e[i].v));
            e[i].v-=w;
            e[i^1].v+=w;
            used+=w;
            if (used==f)return f;
        }
    if (!used)h[x]=-1;
    return used;
}
inline void dinic()
{while (bfs())ans+=dfs(S,inf);}
int main()
{
    n=read();m=read();
    for (int i=1;i<=m;i++)
    {
        int x=read(),y=read();
        go[x][y]=1;
    }
    for (int k=1;k<=n;k++)
    for (int i=1;i<=n;i++)
    for (int j=1;j<=n;j++)
    if (go[i][k]&&go[k][j])go[i][j]=1;
     
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    if (go[i][j])insert(i,j+n,1);
     
    for (int i=1;i<=n;i++)
    insert(S,i,1),insert(i+n,T,1);
    dinic();
    printf("%d\n",n-ans);
}