现代豪宅

现代豪宅

题解

我们可以将每个有开关的点与起点终点离散化，再分别将每个点横向与竖向建成虚点，表示方向的转换，之间连接距离为1，把每行与每列的点用之间的差值作距离连接，建成一张图，跑一遍dijkstra即可。

源码

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>	
#include<map>
#define pii pair<int,int>
#define MAXN 3000005
#define MAXM 40000006
using namespace std;
typedef long long LL;
#define int LL
#define re register
const int INF=0x7f7f7f7f7f7f;
int m,n,k,cnt;
int to[MAXM],nxt[MAXM],paid[MAXM];
int tot,head[MAXN];
int dis[MAXN];
struct ming
{
	int zb,id;
};
vector<ming> h[MAXN],l[MAXN];
struct node
{
	int x,y;
	bool friend operator < (node x,node y)
	{
		return x.y>y.y;
	}
};
priority_queue<node> q;
bool cmp(ming x,ming y)
{
	return x.zb<y.zb;
}
void addEdge(int u,int v,int w)
{
	to[++tot]=v;paid[tot]=w;
	nxt[tot]=head[u];head[u]=tot;
}
#define gc() getchar()
template<typename _T>
inline void read(_T &x)
{
    _T f=1;x=0;char s=gc();
    while(s>'9'||s<'0'){if(s=='-')f=-1;s=gc();}
    while(s>='0'&&s<='9'){x=(x<<3)+(x<<1)+(s^48);s=gc();}
    x*=f;
}
void dijk()
{
	for(re int i=0;i<=(k<<1|1);i++)
		dis[i]=INF;
	dis[0]=0;
	while(!q.empty()) q.pop();
	q.push((node){0,dis[0]});
	while(!q.empty())
	{
		node t=q.top();q.pop();
		if(t.y>dis[t.x]) continue;
		for(re int i=head[t.x];i;i=nxt[i])
		{
			int v=to[i],w=paid[i];
			if(dis[v]>dis[t.x]+w)
			{
				dis[v]=dis[t.x]+w;
				q.push((node){v,dis[v]});
			}
		}
	}
}
signed main()
{
	read(m);read(n);read(k);
	for(re int i=1;i<=k;i++)
	{
		int x,y;++cnt;
		read(x);read(y);
		l[x].push_back((ming){y,cnt});
		h[y].push_back((ming){x,cnt+k});
	}
	for(re int i=1;i<=n;i++)
	{
		sort(h[i].begin(),h[i].end(),cmp);
		int siz=h[i].size();
		for(re int j=1;j<siz;j++)
		{
			addEdge(h[i][j-1].id,h[i][j].id,h[i][j].zb-h[i][j-1].zb);
			addEdge(h[i][j].id,h[i][j-1].id,h[i][j].zb-h[i][j-1].zb);
		}
	}
	if(h[n].size())
	{
		int si=h[n].size()-1;
		addEdge(k<<1|1,h[n][si].id,m-h[n][si].zb);
		addEdge(h[n][si].id,k<<1|1,m-h[n][si].zb);
	}
	for(re int i=1;i<=m;i++)
	{
		sort(l[i].begin(),l[i].end(),cmp);
		int siz=l[i].size();
		for(re int j=1;j<l[i].size();j++)
		{
			addEdge(l[i][j-1].id,l[i][j].id,l[i][j].zb-l[i][j-1].zb);
			addEdge(l[i][j].id,l[i][j-1].id,l[i][j].zb-l[i][j-1].zb);
		}
	}
	if(l[m].size())
	{
		int si=l[m].size()-1;
		addEdge(k<<1|1,l[m][si].id,n-l[m][si].zb);
		addEdge(l[m][si].id,k<<1|1,n-l[m][si].zb);
	}
	if(l[1].size())
	{
		addEdge(0,l[1][0].id,l[1][0].zb-1);
		addEdge(l[1][0].id,0,l[1][0].zb-1);
	}
	for(re int i=1;i<=k;i++)
	{
		addEdge(i,i+k,1);
		addEdge(i+k,i,1);
	}
	dijk();
	if(dis[k<<1|1]>INF-1)
		puts("-1");
	else printf("%lld",dis[k<<1|1]);
    return 0;
}

谢谢！！！