Description
Dr. Octopus kidnapped Spiderman's girlfriend M.J. and kept her in the West Tower. Now the
hero, Spiderman, has to reach the tower as soon as he can to rescue her, using his own
weapon, the web.
From Spiderman's apartment, where he starts, to the tower there is a straight road.
Alongside of the road stand many tall buildings, which are definitely taller or equal to
his apartment. Spiderman can shoot his web to the top of any building between the tower and
himself (including the tower), and then swing to the other side of the building. At the
moment he finishes the swing, he can shoot his web to another building and make another
swing until he gets to the west tower. Figure-1 shows how Spiderman gets to the tower from
the top of his apartment – he swings from A to B, from B to C, and from C to the tower. All
the buildings (including the tower) are treated as straight lines, and during his swings he
can't hit the ground, which means the length of the web is shorter or equal to the height
of the building. Notice that during Spiderman's swings, he can never go backwards.
You may assume that each swing takes a unit of time. As in Figure-1, Spiderman used 3
swings to reach the tower, and you can easily find out that there is no better way.
The first line of the input contains the number of test cases K (1 <= K <= 20). Each case
starts with a line containing a single integer N (2 <= N <= 5000), the number of buildings
(including the apartment and the tower). N lines follow and each line contains two integers
Xi, Yi, (0 <= Xi, Yi <= 1000000) the position and height of the building. The first
building is always the apartment and the last one is always the tower. The input is sorted
by Xi value in ascending order and no two buildings have the same X value.
Output
For each test case, output one line containing the minimum number of swings (if it
possible to reach the tower) or -1 if Spiderman can
Examples
2
6
0 3
3 5
4 3
5 5
7 4
10 4
3
0 3
3 4
10 4
Output
3
-1
Problem Description
在一维坐标系内，给你n个建筑物的横坐标x，和高度h（保证所有建筑物的高度不小于第一座）。
蜘蛛侠能够在建筑物之间游荡。求从第一座建筑物开始到最后一座的最小游荡次数。如果不能到达则输出-1.
只能以建筑物最高点为圆心进行游荡。
Solution
基础dp
概况：枚举每一座建筑i，再枚举j(j<i),j能通过i所能到达的最远距离是否大于终点的横坐标，如果不大于，
则将其次数加1.
设dp[i]为到达第i个建筑物的最小游荡次数为dp[i].
设第i座建筑物的坐标为(x,y).
则p应满足以下条件：
>第i座建筑物最高点到(p,h)的距离不能大于自己的高度，否则就落地了。>(x−p)2+(y−h)2≤y2>(x−p)2≤y2−(y−h)2>x−p≤y2−(y−h)2−−−−−−−−−−−√>p≥x−y2−(y−h)2−−−−−−−−−−−√>其中p为能够到达第i座建筑物的最远点的横坐标。>h为第一座建筑物的高度。>有对称性可知，每次游荡的起始高度一定为h。>
枚举p到x的所有建筑物的横坐标j，如果x-j+x大于最后一座建筑物的横坐标，则能够到达终点。更新答案
ans=min(ans,dp[j]+1);
否则 dp[x-j+x]=min(dp[x-j+x],dp[j]+1).(x-j+x几何意义即为j关于x的对称点的横坐标)
Code
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 1000005
vector<pair<int,int> >a;
int dp[maxn];
Int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
a.clear();
memset(dp,INF,sizeof(dp));
int n;
cin>>n;
int x,y;
cin>>x>>y;
a.push_back(make_pair(0,0));
for(int i=1;i<n;i++)
{
int u,v;
cin>>u>>v;
a.push_back(make_pair(u,v));
}
dp[0]=0;
int ans=INF;
for(int i=1;i<n;i++)
{
int tx=a[i].first,ty=a[i].second;
for(int j=max(x,tx-(int)sqrt(1.*ty*ty-1.*(ty-y)*(ty-y)));j<tx;j++)
{
if(dp[j]==INF) continue;
int tmp=tx-j+tx;
if(tmp>=a[n-1].first)
{
ans=min(ans,dp[j]+1);
}
else dp[tmp]=min(dp[tmp],dp[j]+1);
}
}
if(ans==INF)
cout<<-1<<endl;
else
cout<<ans<<endl;
}
cin.get(),cin.get();
return 0;
}