[SHOI2003] 吃豆豆

\(Solution\)

考虑把一个豆豆看成一条边，那限制流量为\(1\)，费用为\(1\)，豆豆即会被吃一次。拆点，两点之间连一条有贡献的边和无贡献的边，可做到让两条路径重合后不会多产生贡献。路线不可以相交可以通过连边的方式解决，但我们发现边数太多了。考虑我们连的一些无用的边，形如一个钝角三角形，明显有边\(AC\)是可以不连的，优化后即可过。

\(Code\)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 4005,inf = 2e9;
int h[N],S,T,n,flow[N],vis[N],q[N * 30],tot = 1,pre[N],up[N],m,dis[N],S2,in[N];

struct nd{
	int x,y;
}a[N];
struct edge{
	int to,nxt,f,z;
}e[1000005];
bool cmp(nd x,nd y)
{
	if (x.x == y.x) return x.y < y.y;
	return x.x < y.x;
}
void add(int x,int y,int f,int z)
{
	e[++tot] = edge{y,h[x],f,z},h[x] = tot;
	e[++tot] = edge{x,h[y],0,-z},h[y] = tot;
}
int spfa()
{
	memset(flow,127,sizeof flow);
	memset(dis,127,sizeof dis);
	memset(vis,0,sizeof vis);
	int head = 0,tail = 1;
	q[1] = S,dis[S] = 0,vis[S] = 1,pre[T] = -1;
	while (head < tail)
	{
		int u = q[++head]; vis[u] = 0;
		for (int i = h[u]; i; i = e[i].nxt)
		{
			int v = e[i].to;
        	if (dis[v] > dis[u] + e[i].z && e[i].f > 0)                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    
			{
				dis[v] = dis[u] + e[i].z,pre[v] = u,up[v] = i;
				flow[v] = min(flow[u],e[i].f);
				if (!vis[v]) q[++tail] = v,vis[v] = 1; 
			}
		}
	}
	return (pre[T] != -1);
}
int main()
{
	scanf("%d",&n);
	S = n * 2 + 1,S2 = S + 1,T = S + 2,add(S,S2,2,0);
	for (int i = 1; i <= n; i++) scanf("%d%d",&a[i].x,&a[i].y);
	sort(a + 1,a + 1 + n,cmp);
	for (int i = 2; i <= n; i++)
		for (int j = i - 1,g = -1; j >= 1; j--)
		{
			if (a[j].y > a[i].y) continue;
			if (a[j].y <= g) continue;
			g = a[j].y,add(n + j,i,2,0),in[i]++;
		}
	for (int i = 1; i <= n; i++) if (!in[i]) add(S2,i,2,0); //不这样会TLE
	for (int i = 1; i <= n; i++) add(i,i + n,1,-1),add(i,i + n,1,0),add(i + n,T,2,0);
	int ansc = 0;
	for (; spfa();)
	{
		int o = T; ansc += flow[T] * dis[T];
		for (; o != S; o = pre[o]) e[up[o]].f -= flow[T],e[up[o] ^ 1].f += flow[T];
	}
	printf("%lld\n",-ansc);
}