• 【POJ 1556】The Doors 判断线段相交+SPFA


    黑书上的一道例题:如果走最短路则会碰到点,除非中间没有障碍。

    这样把能一步走到的点两两连边,然后跑SPFA即可。

    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define N 100003
    using namespace std;
    struct Point {
    	double x, y;
    	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
    };
    inline int dcmp(double a) {
    	return (fabs(a) < 1e-6) ? 0 : (a < 0 ? -1 : 1);
    }
    Point operator - (Point a, Point b) {
    	return Point(a.x - b.x, a.y - b.y);
    }
    bool operator == (Point a, Point b) {
    	return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
    }
    inline double Cross(Point a, Point b) {
    	return a.x * b.y - a.y * b.x;
    }
    inline double Dot(Point a, Point b) {
    	return a.x * b.x + a.y * b.y;
    }
    inline bool jiao(Point d1, Point d2, Point d3, Point d4) {
    	return (dcmp(Cross(d4 - d3, d1 - d3)) ^ dcmp(Cross(d4 - d3, d2 - d3))) == -2 &&
    			(dcmp(Cross(d2 - d1, d3 - d1)) ^ dcmp(Cross(d2 - d1, d4 - d1))) == -2;
    }
    inline int bjiao(Point d1, Point d2, Point d3) {
    	if (d1 == d2 || d1 == d3)
    		return 1;
    	if (dcmp(Cross(d2 - d1, d3 - d1)) == 0 && dcmp(Dot(d2 - d1, d3 - d1)) == -1)
    		return 1;
    	return 0;
    }
    inline double dis(Point d1, Point d2) {
    	return sqrt((d2.y - d1.y) * (d2.y - d1.y) + (d2.x - d1.x) * (d2.x - d1.x));
    }
    
    struct nodeline {
    	Point a, b;
    	nodeline(Point _a = (0, 0) , Point _b = (0, 0)) : a(_a), b(_b) {}
    } line[N];
    
    struct node {
    	int nxt, to;
    	double w;
    } E[N << 1];
    
    Point a[N];
    int n, cnt, lcnt, point[N], dd, q[N];
    double nowx, nowy, nowy2, dist[N];
    bool vis[N];
    
    inline void ins(int x, int y, double z) {++dd; E[dd].nxt = point[x]; E[dd].to = y; E[dd].w = z; point[x] = dd;}
    inline double spfa(int S, int T) {
    	memset(dist, 127, sizeof(dist));
    	memset(vis, 0, sizeof(vis));
    	int head = 0, tail = 1;
    	q[1] = S;
    	vis[S] = 1;
    	dist[S] = 0;
    	while (head != tail) {
    		++head; if (head >= N) head %= N;
    		int u = q[head];
    		vis[u] = 0;
    		for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) {
    			int v = E[tmp].to;
    			if (dist[u] + E[tmp].w < dist[v]) {
    				dist[v] = dist[u] + E[tmp].w;
    				if (!vis[v]) {
    					++tail; if (tail >= N) tail %= N;
    					q[tail] = v;
    					vis[v] = 1;
    				}
    			}
    		}
    	}
    	return dist[T];
    }
    
    inline bool check(Point d1, Point d2) {
    	for(int i = 1; i <= lcnt; ++i)
    		if (jiao(d1, d2, line[i].a, line[i].b))
    			return 0;
    	return 1;
    }
    
    int main() {
    	scanf("%d", &n);
    	while (n != -1) {
    		cnt = 2;
    		lcnt = 0;
    		a[1].x = 0;
    		a[1].y = 5;
    		a[2].x = 10;
    		a[2].y = 5;
    		for(int i = 1; i <= n; ++i) {
    			scanf("%lf", &nowx);
    			scanf("%lf", &nowy);
    			a[++cnt] = Point(nowx, nowy);
    			line[++lcnt] = nodeline(Point(nowx, 0), a[cnt]);
    			scanf("%lf", &nowy);
    			a[++cnt] = Point(nowx, nowy);
    			scanf("%lf", &nowy2);
    			a[++cnt] = Point(nowx, nowy2);
    			line[++lcnt] = nodeline(a[cnt - 1], a[cnt]);
    			scanf("%lf", &nowy);
    			a[++cnt] = Point(nowx, nowy);
    			line[++lcnt] = nodeline(a[cnt], Point(nowx, 10));
    		}
    		
    		memset(point, 0 , sizeof(point));
    		dd = 0;
    		for(int i = 1; i <= cnt; ++i)
    			for(int j = i + 1; j <= cnt; ++j)
    				if (check(a[i], a[j]))
    					ins(i, j, dis(a[i], a[j]));
    		for(int i = 1; i <= cnt; ++i)
    			if (i != 2 && check(a[i], a[2]))
    				ins(i, 2, dis(a[i], a[2]));
    		printf("%.2lf
    ", spfa(1,2));
    		scanf("%d", &n);
    	}
    	return 0;
    }
    

    清明节机房也放假啊滚来滚去……~(~o ̄▽ ̄)~o 。。。滚来滚去……o~(_△_o~) ~。。。

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  • 原文地址:https://www.cnblogs.com/abclzr/p/5345620.html
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