pta 编程题16 Saving James Bond - Easy Version

时间:2023-03-09 05:34:37
pta 编程题16 Saving James Bond - Easy Version

其它pta数据结构编程题请参见:pta

题目

主要用到了深度优先搜索。

 #include <iostream>
using namespace std; struct Vertex
{
int x;
int y;
bool marked;
}G[]; int N; //总鳄鱼数
int D; //可以跳的距离
bool dfs(Vertex& v);
bool firstJump(Vertex v);
bool jump(Vertex v1, Vertex v2);
bool success(Vertex v); //可以跳到岸上 int main()
{
int i;
cin >> N >> D;
for (i = ; i < N; i++)
cin >> G[i].x >> G[i].y; bool canEscape = false;
for (i = ; i < N; i++)
{
if (!G[i].marked && firstJump(G[i]))
canEscape = dfs(G[i]);
if (canEscape) break;
}
if (canEscape) cout << "Yes";
else cout << "No";
return ;
} bool firstJump(Vertex v)
{
return v.x * v.x + v.y * v.y <= ( + D) * ( + D);
} bool dfs(Vertex& v)
{
bool canEscape = false;
v.marked = true;
if (success(v)) return true;
for (int i = ; i < N; i++)
{
if (!G[i].marked && jump(v, G[i]))
canEscape = dfs(G[i]);
if (canEscape) break;
}
return canEscape;
} bool success(Vertex v)
{
if (v.x <= D - || v.x >= - D || v.y <= D - || v.y >= - D)
return true;
return false;
} bool jump(Vertex v1, Vertex v2)
{
return (v1.x - v2.x) * (v1.x - v2.x) + (v1.y - v2.y) * (v1.y - v2.y) <= D*D;
}