卧薪尝胆,厚积薄发。
CTSC2008 祭祀
Date: Mon Jul 02 19:43:57 CST 2018 In Category: NoCategory

Description:

给出一张有向无环图,问最多能选几个点使得一个点不能到达另一个点。
$1\le N \le 100$

Solution:

题目要求最长反链长度,由 $Dilworth$ 定理得最长反链 $=$ 最小链覆盖,于是先用 $floyed$ 传递闭包再做拆点二分图匹配用 $n-ans$ 即可。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int n,m;
#define MAXN 100
#define MAXM 1000
bool g[MAXN << 1][MAXN << 1];
struct edge
{
int to,nxt;
}e[MAXN * MAXN * 16];
int edgenum = 0;
int lin[MAXN << 3] = {0};
void add(int a,int b)
{
++edgenum;e[edgenum].to = b;e[edgenum].nxt = lin[a];lin[a] = edgenum;
++edgenum;e[edgenum].to = a;e[edgenum].nxt = lin[b];lin[b] = edgenum;
return;
}
bool v[MAXN << 2];
int match[MAXN << 2];
bool find(int k)
{
for(int i = lin[k];i != 0;i = e[i].nxt)
{
if(!v[e[i].to])
{
v[e[i].to] = true;
if(match[e[i].to] == -1 || find(match[e[i].to]))
{
match[e[i].to] = k;
return true;
}
}
}
return false;
}
int main()
{
scanf("%d%d",&n,&m);
int a,b;
memset(g,0,sizeof(g));
for(int i = 1;i<= m;++i)
{
scanf("%d%d",&a,&b);
g[a][b] = true;
}
for(int k = 1;k <= n;++k)
{
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= n;++j)
{
g[i][j] |= (g[i][k] & g[k][j]);
}
}
}
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= n;++j)
{
if(g[i][j])add(i,j + n);
}
}
memset(match,-1,sizeof(match));
int ans = 0;
for(int i = 1;i <= n;++i)
{
memset(v,0,sizeof(v));
if(find(i))++ans;
}
cout << (n - ans) << endl;
return 0;
}
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