UVALive4374 Drive through MegaCity

题目戳这里。

首先我们对坐标进行离散化，有用的点就变成了(O(N))个。我们假设(A)点(B)的右边(从(A)往(B)跑和从(B)往(A)跑等价)，然后我们很容易发现不会往左跑。于是我们就可以dp了。我们用(f[i][j])表示(A)到((i,j))的最小代价(((i,j))是离散后的坐标)，然后我们很容易得到dp方程。

[f[i][j] = min { f[i-1][j]+Tx[i-1][j] imes (x[i]-x[i-1]),\f[i][j+1]+Ty[i][j] imes (y[j+1]-y[j]),f[i][j-1]+Ty[i][j-1]+(y[j]-y[j-1]) } ]

其中(Tx[i][j])表示从((i,j))走到((i+1,j))走一个单位所需要的时间，(Ty[i][j])表示从((i,j))走到((i,j+1))一个单位所需要的时间。这个我们可以预处理出来(大致就是看这中间有没有点在矩形的边界或中间)。

然后若(B)在(A)下方，则还可能只向下向左向右走。我们可以旋转坐标，就跟上面的一样处理了。

#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

typedef long long ll;
const int maxn = 2010; const ll inf = 1LL<<60;
int Xa,Ya,Xb,Yb,N,totx,toty,bacx[maxn],bacy[maxn],in[maxn][maxn],Lb[maxn][maxn],Db[maxn][maxn];
ll f[maxn][maxn],Tx[maxn][maxn],Ty[maxn][maxn],ans;

inline int gi()
{
	char ch; int ret = 0,f = 1;
	do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
	if (ch == '-') f = -1,ch = getchar();
	do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
	return ret*f;
}

struct Node
{
	int X1,Y1,X2,Y2,T;
	inline void read()
	{
		bacx[++totx] = X1 = gi(); bacy[++toty] = Y1 = gi();
		bacx[++totx] = X2 = gi(); bacy[++toty] = Y2 = gi();
		T = gi();
	}
	inline void convert() { swap(X1,Y1); swap(X2,Y2); }
}rec[maxn];

inline int find(int arr[],int r,int key)
{
	int l = 1,mid;
	while (l <= r)
	{
		mid = (l+r) >> 1;
		if (arr[mid] < key) l = mid+1;
		else r = mid-1;
	}
	return l;
}
	
inline void work(int xx[],int tx,int yy[],int ty)
{
	for (int i = 1;i <= tx;++i) for (int j = 1;j <= ty;++j) Lb[i][j] = Db[i][j] = in[i][j] = 0;
	for (int i = 1;i <= N;++i)
	{
		int pX1 = find(xx,tx,rec[i].X1),pX2 = find(xx,tx,rec[i].X2);
		int pY1 = find(yy,ty,rec[i].Y1),pY2 = find(yy,ty,rec[i].Y2);
		for (int j = pX1+1;j < pX2;++j)
			for (int k = pY1+1;k < pY2;++k) in[j][k] = i;
		for (int k = pY1+1;k < pY2;++k) Lb[pX1][k] = i;
		for (int j = pX1+1;j < pX2;++j) Db[j][pY1] = i;
	}
	for (int i = 1;i <= tx;++i)
		for (int j = 1;j <= ty;++j)
		{
			if (i < tx)
			{
				Tx[i][j] = 10;
				if (in[i][j]) Tx[i][j] = rec[in[i][j]].T;
				else if (in[i+1][j]) Tx[i][j] = rec[in[i+1][j]].T;
				else if (Lb[i][j]) Tx[i][j] = rec[Lb[i][j]].T;
			}
			if (j < ty)
			{
				Ty[i][j] = 10;
				if (in[i][j]) Ty[i][j] = rec[in[i][j]].T;
				else if (in[i][j+1]) Ty[i][j] = rec[in[i][j+1]].T;
				else if (Db[i][j]) Ty[i][j] = rec[Db[i][j]].T;
			}
		}
	if (Xa > Xb) swap(Xa,Xb),swap(Ya,Yb);
	
	for (int i = 1;i <= tx;++i) for (int j = 1;j <= ty;++j) f[i][j] = inf;
	f[find(xx,tx,Xa)][find(yy,ty,Ya)] = 0;
	for (int i = 2;i <= tx;++i)
	{
		for (int j = 1;j <= ty;++j)
		{
			f[i][j] = min(f[i-1][j]+Tx[i-1][j]*(xx[i]-xx[i-1]),f[i][j]);
			if (j > 1) f[i][j] = min(f[i][j],f[i][j-1]+Ty[i][j-1]*(yy[j]-yy[j-1]));
		}
		for (int j = ty;j;--j) if (j < ty) f[i][j] = min(f[i][j],f[i][j+1]+Ty[i][j]*(yy[j+1]-yy[j]));
	}
	ans = min(ans,f[find(xx,tx,Xb)][find(yy,ty,Yb)]);
}

int main()
{
	freopen("4374.in","r",stdin);
	freopen("4374.out","w",stdout);
	while (scanf("%d %d %d %d",&Xa,&Ya,&Xb,&Yb) != EOF)
	{
		totx = toty = 0; N = gi(); ans = inf;
		bacx[++totx] = Xa; bacx[++totx] = Xb;
		bacy[++toty] = Ya; bacy[++toty] = Yb;
		for (int i = 1;i <= N;++i) rec[i].read();

		sort(bacx+1,bacx+totx+1); sort(bacy+1,bacy+toty+1);
		totx = unique(bacx+1,bacx+totx+1)-bacx-1; toty = unique(bacy+1,bacy+toty+1)-bacy-1;
	
		work(bacx,totx,bacy,toty);

		swap(Xa,Ya); swap(Xb,Yb);
		for (int i = 1;i <= N;++i) rec[i].convert();
		work(bacy,toty,bacx,totx);
		cout << ans << endl;		
	}
	fclose(stdin); fclose(stdout);
	return 0;
}