Topic links: http://acm.hdu.edu.cn/showproblem.php?pid=6670
Mindis
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 548 Accepted Submission(s): 119
Problem Description
has n rectangles, the rectangle sides parallel to the coordinate axes on a plane, is now a need to control the degree of the character come to bear from A to B point.
The character can move up and down, just in the area covered by the rectangular k, the rate is the role distances k + 1 / sec (inclusive rectangular footprint area).
A request to move to the fastest B how many seconds.
Input
first line an integer T (1≤T≤5) represents the number of data sets.
For each test, a first line of the input integer n (1≤n≤200).
The next four lines each n integers x1, y1, x2, y2 ( 0≤x1 <x2≤1000000000,0≤y1 <y2≤1000000000), represent the coordinates of the lower left and upper right corner of the rectangle.
The last line four integers xa, ya, xb, yb ( (0≤xa, xb, ya, yb≤1000000000) representative of the coordinates of A and B.
Output
For each test, the output of a decimal answer. The answer to five decimal places.
Sample Input
1
1
5 5 6 6
7 7 8 8
Sample Output
2.00000
Source
2019 Nian Baidu Star · Programming Contest - a preliminary round
Ideas:
First of all topics given points at discrete, give a n * m grid, n and m are the largest 400
We enumerate each rectangle, after which point the coordinates of the rectangle while the enumeration of discrete included,
This can maintain a number of points of each rectangle is included,
cnt [x] [y] [k] is represented on the abscissa grid x, ordinate y, the direction is k, (0,1,2,3 representative of vertical and horizontal) of this discretization length of the sides 1 the number of times a rectangular footprint.
Then built adjacent dots (the edge distance cost / speed, i.e. the time obtained by subtracting the distance we can coordinate, the speed is derived from the number of rectangles includes the point), the shortest path map can be run,
Write code is very long, it is not difficult to write, because writing does not require a lot of mental and physical part is mechanical operation. , Attention to detail,
See details Code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 400010;
const ll inf = 1e18;
/*** TEMPLATE CODE * * STARTS HERE ***/
int t;
int n;
struct juzhen
{
int x1,x2,y1,y2;
}a[500];
std::vector<int> vx,vy;
int sn,sm;
int getid(int x,int y)
{
return (x-1)*sn+y;
}
double dis[maxn];
struct node
{
int to;
double val;
node (){}
node(int tt,double vv)
{
to=tt;
val=vv;
}
bool operator <(const node &b)const
{
return val>b.val;
}
};
std::vector<node> v[maxn];
int cnt[500][500][5];
bool check(int x,int y)
{
return x>=1&&x<=sn&&y>=1&&y<=sm;
}
juzhen aid;
priority_queue<node> heap;
void dij()
{
while(heap.size())
{
heap.pop();
}
dis[getid(aid.x1,aid.y1)]=0;
heap.push(node(getid(aid.x1,aid.y1),0));
node temp;
while(!heap.empty())
{
temp=heap.top();
heap.pop();
for(auto x:v[temp.to])
{
if(dis[x.to]>x.val+dis[temp.to])
{
dis[x.to]=x.val+dis[temp.to];
heap.push(node(x.to,dis[x.to]));
}
}
}
}
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
gbtb;
cin>>t;
while(t--)
{
cin>>n;
vx.clear();
vy.clear();
repd(i,1,n)
{
cin>>a[i].x1>>a[i].y1>>a[i].x2>>a[i].y2;
vx.push_back(a[i].x1);
vx.push_back(a[i].x2);
vy.push_back(a[i].y1);
vy.push_back(a[i].y2);
}
cin>>aid.x1>>aid.y1>>aid.x2>>aid.y2;
vx.push_back(aid.x1);
vx.push_back(aid.x2);
vy.push_back(aid.y1);
vy.push_back(aid.y2);
sort(ALL(vx));
sort(ALL(vy));
vx.erase(unique(ALL(vx)),vx.end());
vy.erase(unique(ALL(vy)),vy.end());
sn=sz(vx);
sm=sz(vy);
repd(i,1,sn)
{
repd(j,1,sm)
{
dis[getid(i,j)]=inf;
v[getid(i,j)].clear();
repd(z,0,3)
cnt[i][j][z]=1;
}
}
aid.x1=lower_bound(ALL(vx),aid.x1)-vx.begin()+1;
aid.x2=lower_bound(ALL(vx),aid.x2)-vx.begin()+1;
aid.y1=lower_bound(ALL(vy),aid.y1)-vy.begin()+1;
aid.y2=lower_bound(ALL(vy),aid.y2)-vy.begin()+1;
repd(i,1,n)
{
int x1=lower_bound(ALL(vx),a[i].x1)-vx.begin()+1;
int y1=lower_bound(ALL(vy),a[i].y1)-vy.begin()+1;
int x2=lower_bound(ALL(vx),a[i].x2)-vx.begin()+1;
int y2=lower_bound(ALL(vy),a[i].y2)-vy.begin()+1;
repd(j,x1+1,x2-1)
{
repd(k,y1+1,y2-1)
{
repd(z,0,3)
cnt[j][k][z]++;
}
}
// 0 1 2 3
// 上 下 左 右
repd(j,x1+1,x2-1)
{
cnt[j][y2][1]++;
cnt[j][y2][2]++;
cnt[j][y2][3]++;
cnt[j][y1][0]++;
cnt[j][y1][2]++;
cnt[j][y1][3]++;
}
repd(j,y1+1,y2-1)
{
cnt[x1][j][0]++;
cnt[x1][j][1]++;
cnt[x1][j][3]++;
cnt[x2][j][0]++;
cnt[x2][j][1]++;
cnt[x2][j][2]++;
}
cnt[x1][y1][0]++;
cnt[x1][y1][3]++;
cnt[x2][y1][0]++;
cnt[x2][y1][2]++;
cnt[x1][y2][3]++;
cnt[x1][y2][1]++;
cnt[x2][y2][1]++;
cnt[x2][y2][2]++;
}
repd(i,1,sn)
{
repd(j,1,sm)
{
if(check(i-1,j))
{
v[getid(i,j)].push_back(node(getid(i-1,j),1.00*(vx[i-1]-vx[i-2])/cnt[i][j][2]));
}
if(check(i+1,j))
{
v[getid(i,j)].push_back(node(getid(i+1,j),1.00*(vx[i]-vx[i-1])/cnt[i][j][3]));
}
if(check(i,j-1))
{
v[getid(i,j)].push_back(node(getid(i,j-1),1.00*(vy[j-1]-vy[j-2])/cnt[i][j][1]));
}
if(check(i,j+1))
{
v[getid(i,j)].push_back(node(getid(i,j+1),1.00*(vy[j]-vy[j-1])/cnt[i][j][0]));
}
}
}
dij();
// cout<<getid(aid.x2,aid.y2)<<endl;
cout<<fixed<<setprecision(5)<<dis[getid(aid.x2,aid.y2)]<<endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}