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大致题意:在二维平面上,给一些圆形岛屿的坐标和半径,以及圆形船的位置和半径,问能否划到无穷远的地方去
思路:考虑任意两点,如果a和b之间船不能通过,则连一条边,则问题转化为判断点是否在多边形中。先进行坐标变换,将船变到原点,以从原点到每个点的有向角作为状态,每条边的边权为这条边对有向角的改变量,那么点在多边形内相当于存在负权环,用SPFA判负环即可。
#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE namespace Debug { void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} } #endif // ONLINE_JUDGE template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 0x3f3f3f3f; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ const int maxn = 3e2 + 7; int dcmp(double x) { if (fabs(x) < EPS) return 0; return x > 0? 1 : - 1; } struct Circle { double x, y, r; Circle(double x, double y, double r) { this->x = x; this->y = y; this->r = r; } void read() { scanf("%lf%lf%lf", &x, &y, &r); } Circle() {} }; Circle p[maxn]; double e[maxn][maxn], d[maxn]; bool vis[maxn]; int n, cnt[maxn]; double dist(double x1, double y1, double x2, double y2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); } bool relax(double u, double w, double &v) { if (dcmp(v - u - w) > 0) { v = u + w; return true; } return false; } bool spfa(int s) { queue<int> Q; Q.push(s); fillchar(vis, 0); for (int i = 0; i < n; i ++) { d[s] = INF; } fillchar(cnt, 0); d[s] = 0; while (!Q.empty()) { int H = Q.front(); Q.pop(); vis[H] = false; for (int i = 0; i < n; i ++) { if (e[H][i] < INF) { if (relax(d[H], e[H][i], d[i]) && !vis[i]) { if (cnt[i] >= n) return true; Q.push(i); vis[i] = true; cnt[i] ++; } } } } return false; } void work() { for (int i = 0; i < n; i ++) { if (spfa(i)) { puts("NO"); return ; } } puts("YES"); } double calcangle(int i, int j) { Circle a = p[i], b = p[j]; double angle = acos((a.x * b.x + a.y * b.y) / dist(a.x, a.y, 0, 0) / dist(b.x, b.y, 0, 0)); if (dcmp(a.x * b.y - a.y * b.x) <= 0) return angle; return - angle; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE while (cin >> n) { for (int i = 0; i < n; i ++) { p[i].read(); } Circle me; me.read(); for (int i = 0; i < n; i ++) { p[i].x -= me.x; p[i].y -= me.y; } for (int i = 0; i < n; i ++) { for (int j = 0; j < n; j ++) { e[i][j] = INF + 1; } } for (int i = 0; i < n; i ++) { for (int j = i + 1; j < n; j ++) { double buf = dist(p[i].x, p[i].y, p[j].x, p[j].y) - p[i].r - p[j].r; if (dcmp(buf - me.r * 2) < 0) { e[i][j] = calcangle(i, j); e[j][i] = - e[i][j]; //Debug::print(i, j, e[i][j]); } } } work(); } return 0; }
[ACdream 1215 Get Out!]判断点在封闭图形内
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原文地址:http://www.cnblogs.com/jklongint/p/4759186.html