思路:
先建双向边:鳄鱼与鳄鱼之间(若距离<=d),圆心与鳄鱼之间(距离-7.5<=d),汇点(n+1)与鳄鱼之间建边(从鳄鱼的坐标能到边界)
跑0~n+1的最短路,因为本题卡精度!!! 所以abs(a-b)<=eps;判相等
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<algorithm>
using namespace std;
const int N = 2e2 + 5;
const double Inf = 1e9;
const double eps = 1e-4;
typedef pair<double, int> P;
vector<P> v[N];
int n;
double x[N], y[N], jump, di;
inline void add(int x, int y, double c) {
v[x].push_back(P(c, y));
}
inline double pow(double x) {
return x * x;
}
inline double getDistance(double x0, double y0, double x1, double y1) {
return sqrt(pow(x0 - x1) + pow(y0 - y1));
}
void Dijstra(int from, int n) {
int vis[N], path[N];
double dis[N];
for(int i = 1; i <= n + 1; i += 1) {
dis[i] = Inf;
vis[i] = false;
path[i] = (int)Inf;
}
priority_queue<P, vector<P>, greater<P> >que;
dis[from] = 0;
path[from] = vis[from] = 0;
que.push(P(0, from));
while(!que.empty()) {
int x = que.top().second;
que.pop();
if(vis[x])
continue;
vis[x] = 1;
for(auto e : v[x]) {
int to = e.second;
double c = e.first;
if(abs(dis[to] - dis[x] - c) <= eps) {
if(path[to] > path[x] + 1)
path[to] = path[x] + 1;
} else if(dis[to] >= dis[x] + c) {
dis[to] = dis[x] + c;
path[to] = path[x] + 1;
que.push(P(dis[to], to));
}
}
}
if(abs(dis[n + 1] - Inf) <= eps)
printf("can't be saved\n");
else
printf("%.2f %d\n", dis[n + 1], path[n + 1]);
}
void BuildEdges() {
v[0].clear();
for(int i = 1; i <= n; i += 1) {
di = sqrt(pow(x[i]) + pow(y[i]));
if(di - 7.50 <= jump)
add(0, i, di - 7.50);
di = min(min(50.0 - x[i], 50.0 - y[i]), min(50.0 + x[i], 50.0 + y[i]));
if(di <= jump)
add(i, n + 1, di);
for(int j = i + 1; j <= n; j += 1) {
di = getDistance(x[i], y[i], x[j], y[j]);
if(di <= jump) {
add(i, j, di);
add(j, i, di);
}
}
}
}
int main() {
x[0] = y[0] = 0;
while(~scanf("%d%lf", &n, &jump)) {
for(int i = 1; i <= n; i += 1) {
scanf("%lf%lf", &x[i], &y[i]);
v[i].clear();
}
if(jump >= 42.50) {
printf("42.50 1\n");
continue;
}
BuildEdges();
Dijstra(0, n);
}
return 0;
}