BZOJ 4823 Luogu P3756 [CQOI2017]老C的方块 (网络流、最小割)

题目链接

(Luogu) https://www.luogu.org/problem/P3756

(BZOJ) http://lydsy.com/JudgeOnline/problem.php?id=4823

题解

有点神仙的最小割题。

考虑题目里的图形，如果我们用四种颜色对棋盘进行染色，奇数行依次染\(0,1,2,3,0,1,2,3...\), 偶数行依次染\(3,2,1,0,3,2,1,0...\)则条件可以转化为不能出现相连的\(4\)个颜色互不相同的块。

那么可以建一个四层的图，对于每条两侧都有关键点的特殊边，按照\(S\rightarrow 3\rightarrow 0\rightarrow 1\rightarrow 2\rightarrow T\)的顺序连边，其中\(S\rightarrow 3\)连\(3\)色点的点权，\(0\rightarrow 1\)连两个关键点权值的最小值，\(2\rightarrow T\)连\(2\)色点的点权。不出现相连的四个颜色互不相同的块等价于不存在从\(S\)到\(T\)的路径。

然后跑最小割即可。

因为是分层图，所以dinic跑得很快(复杂度应该是\(O(n\sqrt n)\))，可以通过此题。

代码

#include<bits/stdc++.h>
#define llong long long
using namespace std;

const int INF = 1e9;
namespace NetFlow
{
	const int N = 1e5+2;
	const int M = 8e5;
	struct Edge
	{
		int v,w,nxt,rev;
	} e[(M<<1)+3];
	int fe[N+3];
	int te[N+3];
	int dep[N+3];
	int que[N+3];
	int n,en,s,t;
	void addedge(int u,int v,int w)
	{
//		printf("addedge %d %d %d\n",u,v,w);
		en++; e[en].v = v; e[en].w = w;
		e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;
		en++; e[en].v = u; e[en].w = 0;
		e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
	}
	bool bfs()
	{
		for(int i=1; i<=n; i++) dep[i] = 0;
		int head = 1,tail = 1; que[1] = s; dep[s] = 1;
		while(head<=tail)
		{
			int u = que[head]; head++;
			for(int i=fe[u]; i; i=e[i].nxt)
			{
				int v = e[i].v;
				if(e[i].w>0 && dep[v]==0)
				{
					dep[v] = dep[u]+1;
					if(v==t)return true;
					tail++; que[tail] = v;
				}
			}
		}
		return false;
	}
	int dfs(int u,int cur)
	{
		if(u==t||cur==0) {return cur;}
		int rst = cur;
		for(int &i=te[u]; i; i=e[i].nxt)
		{
			int v = e[i].v;
			if(e[i].w>0 && rst>0 && dep[v]==dep[u]+1)
			{
				int flow = dfs(v,min(rst,e[i].w));
				if(flow>0)
				{
					e[i].w -= flow;
					rst -= flow;
					e[e[i].rev].w += flow;
					if(rst==0) {return cur;}
				}
			}
		}
		if(rst==cur) {dep[u] = -2;}
		return cur-rst;
	}
	int dinic(int _n,int _s,int _t)
	{
		n = _n,s = _s,t = _t;
		int ret = 0;
		while(bfs())
		{
			for(int i=1; i<=n; i++) te[i] = fe[i];
			memcpy(te,fe,sizeof(int)*(n+1));
			ret += dfs(s,INF);
		}
		return ret;
	}
}
using NetFlow::addedge;
using NetFlow::dinic;

const int N = 1e5;
struct Point
{
	int x,y,w;
} a[N+3];
map<int,int> mp[N+3];
int id[N+3],clr[N+3];
int n,nx,ny;

int getclr(int x,int y)
{
	if(y&1) {return (x-1)&3;}
	else {return 3-((x-1)&3);}
}

int main()
{
	scanf("%d%d%d",&nx,&ny,&n);
	for(int i=1; i<=n; i++)
	{
		scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].w);
		mp[a[i].x][a[i].y] = i;
	}
	for(int i=1; i<=n; i++)
	{
		int x = a[i].x,y = a[i].y,clr = getclr(x,y);
		if(((x+(y<<1))&3)==3 && mp[x+1].count(y))
		{
			int j = mp[x+1][y],w = min(a[i].w,a[j].w);
			if(clr==0)
			{
				int k = mp[x-1][y]; if(k) {addedge(k+2,i+2,INF);}
				k = mp[x][y-1]; if(k) {addedge(k+2,i+2,INF);}
				k = mp[x][y+1]; if(k) {addedge(k+2,i+2,INF);}
				k = mp[x+2][y]; if(k) {addedge(j+2,k+2,INF);}
				k = mp[x+1][y+1]; if(k) {addedge(j+2,k+2,INF);}
				k = mp[x+1][y-1]; if(k) {addedge(j+2,k+2,INF);}
				addedge(i+2,j+2,w);
			}
			else if(clr==1)
			{
				int k = mp[x+2][y]; if(k) {addedge(k+2,j+2,INF);}
				k = mp[x+1][y+1]; if(k) {addedge(k+2,j+2,INF);}
				k = mp[x+1][y-1]; if(k) {addedge(k+2,j+2,INF);}
				k = mp[x-1][y]; if(k) {addedge(i+2,k+2,INF);}
				k = mp[x][y-1]; if(k) {addedge(i+2,k+2,INF);}
				k = mp[x][y+1]; if(k) {addedge(i+2,k+2,INF);}
				addedge(j+2,i+2,w);
			}
		}
		else if(clr==3)
		{
			addedge(1,i+2,a[i].w);
		}
		else if(clr==2)
		{
			addedge(i+2,2,a[i].w);
		}
	}
	int ans = dinic(n+2,1,2);
	printf("%d\n",ans);
	return 0;
}