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[ACdream 1215 Get Out!]判断点在封闭图形内

时间:2015-08-26 07:04:35      阅读:213      评论:0      收藏:0      [点我收藏+]

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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

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