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Question 249. 2022 Winter Holiday Ladder Training-7-13 Rescue 007 (Upgrade) (30 points)
1. The topic
2. Problem solving
Before looking at the code of this question, you can take a look at the previous non-upgraded version of Rescue 007. After all, the ideas are basically the same, but here are some condiments for the results to be output. code show as below:
#include <bits/stdc++.h>
using namespace std;
const int maxn=101;
int N;
double D;
int vis[maxn];
int parent[maxn];//parent用于表示跳到i号鳄鱼之前的那只鳄鱼的编号，简称父鳄鱼
vector<int> last;//用于存储所有的可达岸边的最终鳄鱼的编号
struct Point
{
double x,y;
int index;//用于存储鳄鱼的下标，这是方便得到parent
} P[maxn];
double getDistance(Point p1,Point p2)//计算两点间距离
{
return sqrt(pow(p1.x-p2.x,2)+pow(p1.y-p2.y,2));
}
int judge(Point p)//判断p鳄鱼能否让007上岸
{
if(p.x-D<=-50||p.x+D>=50||p.y-D<=-50||p.y+D>=50)
{
return 1;
}
else
{
return 0;
}
}
int bfs(struct Point p0)
{
if(judge(p0))//看能否从岛一步到岸上—>最后一个测试点。。。
{
return 1;
}
queue<Point> q;
q.push(p0);
while(!q.empty())
{
Point tmp=q.front();
q.pop();
if(tmp.x==0&&tmp.y==0)
{
for(int i=0; i<N; i++)
{
if(getDistance(tmp,P[i])<=D+7.5&&vis[i]!=1)
{
q.push(P[i]);
vis[i]=1;
parent[i]=tmp.index;//设置i的父鳄鱼为tmp鳄鱼的编号
if(judge(P[i]))
{
last.push_back(i);//通过鳄鱼i可以跳到岸上，i放入last
}
}
}
}
else
{
for(int i=0; i<N; i++)
{
if(getDistance(tmp,P[i])<=D&&vis[i]!=1)
{
q.push(P[i]);
vis[i]=1;
parent[i]=tmp.index;
if(judge(P[i]))
{
last.push_back(i);
}
}
}
}
}
return last.size();
}
int main()
{
cin.tie(0),cout.tie(0);
cin>>N>>D;
for(int i=0; i<N; i++)
{
cin>>P[i].x>>P[i].y;
P[i].index=i;
}
struct Point p0;
p0.x=0,p0.y=0,p0.index=-1;
if(bfs(p0))
{
stack<Point> res;//最终鳄鱼路径
for(int i=0; i<last.size(); i++)
{
stack<Point> s;//存储以last为终点鳄鱼的走过的鳄鱼路径
while(parent[last[i]]!=-1)//往回一直到根鳄鱼，以寻迹
{
s.push(P[last[i]]);
last[i]=parent[last[i]];
}
s.push(P[last[i]]);
if(res.size()==0||res.size()>s.size())//选取步数最小，即鳄鱼数最少
{
res=s;
}
else if(res.size()==s.size()&&getDistance(p0,s.top())<getDistance(p0,res.top()))//步数相同则考虑到第一只鳄鱼最小者
{
res=s;
}
}
cout<<res.size()+1<<endl;//步数等于鳄鱼数+1
while(!res.empty())
{
cout<<res.top().x<<" "<<res.top().y<<endl;
res.pop();
}
}
else
{
cout<<0<<endl;
}
}