原题链接
题意：给定起点和终点还有一些线段，要求不能穿越线段从起点到终点的最短路径。
注意线段的端点是可以选的，而且可以沿着线段走。
考虑到k的范围只有300，因此考虑到n^3的枚举，我们可以枚举每两个点，然后再根据k条线判断是否相交，如果都不相交，那么就可以在两点之间连一条线了。
至于判断是否相交，我们可以用向量叉积的方法，详细证明就不多说了。
#include <bits/stdc++.h>
using namespace std;
#define ACM_LOCAL
typedef long long ll;
const ll MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
const double eps = 1e-6;
int stx, sty, edx, edy;
int n, m, k, idx;
struct Point {
double x,y;
Point(){
}
Point(double x,double y) {
this -> x = x;
this -> y = y;
}
Point operator-(Point a)
{
return Point(x - a.x,y - a.y);
}
double operator*(Point a)
{
return x * a.y - y * a.x;
}
};
double get_dis(Point p1,Point p2) {
return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}
struct Line {
Point a,b;
}l[N];
bool judge(Line line1, Line line2) {
Point AD = line1.a - line2.a;
Point AC = line1.a - line2.b;
Point AB = line1.a - line1.b;
Point CA = line2.a - line1.a;
Point CB = line2.a - line1.b;
Point CD = line2.a - line2.b;
if((CD * CA) * (CD * CB) < 0 && (AB * AC) * (AB * AD) < 0)
return true;
return false;
}
struct edge {
int to, next;
double w;
}e[N<<1];
int cnt, h[N], vis[N];
double dis[N];
void add(int u, int v, double w) {
e[cnt].to = v;
e[cnt].next = h[u];
e[cnt].w = w;
h[u] = cnt++;
}
struct node {
int id;
double dis;
bool operator < (const node &rhs) const {
return dis > rhs.dis;
}
};
void dij(int st, int ed) {
priority_queue<node> q;
for (int i = 0; i <= ed; i++) dis[i] = INF;
dis[st] = 0;
q.push({
st, 0});
while (q.size()) {
int u = q.top().id;
q.pop();
if (vis[u]) continue;
vis[u] = 1;
for (int i = h[u]; ~i; i = e[i].next) {
int v = e[i].to;
if (dis[v] > dis[u] + e[i].w) {
dis[v] = dis[u] + e[i].w;
if (!vis[v]) {
q.push({
v, dis[v]});
}
}
}
}
}
void solve() {
vector<Point> vcc;
vcc.push_back({
0, 0});
scanf("%d %d %d", &n, &m, &k);
memset(h, -1, sizeof h);
for (int i = 1; i <= k; i++) {
Point a1, a2;
scanf("%lf%lf", &a1.x, &a1.y);
scanf("%lf%lf", &a2.x, &a2.y);
l[i].a = a1;
l[i].b = a2;
vcc.push_back(a1);
vcc.push_back(a2);
add(2*i-1, 2*i, get_dis(a1, a2));
add(2*i-1, 2*i, get_dis(a1, a2));
}
Point st, ed;
scanf("%lf%lf", &st.x, &st.y);
scanf("%lf%lf", &ed.x, &ed.y);
vcc.push_back(st);
vcc.push_back(ed);
for (int i = 1; i <= k*2+2; i++) {
for (int j = i+1; j <= k*2+2; j++) {
Line line1 = {
vcc[i], vcc[j]};
int f = 0;
for (int o = 1; o <= k; o++) {
if (judge(line1, l[o])) {
f = 1;
break;
}
}
if (!f) add(i, j, get_dis(vcc[i], vcc[j])), add(j, i, get_dis(vcc[i], vcc[j]));
}
}
dij(k*2+1, k*2+2);
printf("%.4lf\n", dis[k*2+2]);
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
solve();
return 0;
}