Gym - 100851F Froggy Ford (Dijkstra)

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题目链接：https://cn.vjudge.net/problem/Gym-100851F

这道题是POJ-2253的升级版：https://blog.csdn.net/GYH0730/article/details/77510915

在POJ2253的基础上要加上一个点，显然这个点要加在两点的中点处，提前预处理出任意一点到S，到T的所有路径上最大边的最小值，然后我们可以枚举所有的边，答案为min(max（ds[u], w(u,v)/2, dt[v] )，记录一下最小值就行了

细节上要注意边界的处理

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1050;
const double INF = 1e18;
struct Edge
{
    int from,to;
    double dist;
    Edge(int u,int v,double d):from(u),to(v),dist(d) {}
};
struct HeapNode
{
    double d;
    int u;
    HeapNode(int _u,double _d):u(_u),d(_d){}
    bool operator < (const HeapNode& rhs) const
    {
        return d > rhs.d;
    }
};
struct Dijkstra
{
    int n,m;
    vector<Edge> edges;
    vector<int> G[MAXN];
    bool vis[MAXN];
    double dis[MAXN];
    void init(int n)
    {
        this -> n = n;
        for(int i = 0; i <= n + 1; i++) G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,double dist)
    {
        edges.push_back(Edge(from,to,dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }
    void dijkstra(int s)
    {
        priority_queue<HeapNode> Q;
        for(int i = 0; i <= n + 1; i++) {
            dis[i] = INF;
            vis[i] = false;
        }
        dis[s] = 0.0;
        Q.push(HeapNode(s,0.0));
        while(!Q.empty()) {
            HeapNode x = Q.top();
            Q.pop();
            int u = x.u;
            if(vis[u]) continue;
            vis[u] = true;
            for(int i = 0; i < G[u].size(); i++) {
                Edge& e = edges[G[u][i]];
                if(dis[e.to] > max(dis[u],e.dist)) {
                    dis[e.to] = max(dis[u],e.dist);
                    Q.push(HeapNode(e.to,dis[e.to]));
                }
            }
        }
    }
}ds,dt;
struct Point
{
    double x,y;
}p[MAXN];
inline double cal(int i,int j)
{
    return sqrt( (p[i].x - p[j].x) * (p[i].x - p[j].x) +
                 (p[i].y - p[j].y) * (p[i].y - p[j].y)
                );
}
int main(void)
{
    freopen("froggy.in","r",stdin);
    freopen("froggy.out","w",stdout);
    double w,temp;
    int n;
    scanf("%lf %d",&w,&n);
    ds.init(n);dt.init(n);
    int S = 0,T = n + 1;
    for(int i = 1; i <= n; i++) {
        scanf("%lf %lf",&p[i].x,&p[i].y);
    }
    for(int i = 1; i <= n; i++) {
        ds.AddEdge(S,i,p[i].x);
        dt.AddEdge(i,S,p[i].x);
        ds.AddEdge(i,T,w - p[i].x);
        dt.AddEdge(T,i,w - p[i].x);
        for(int j = i + 1; j <= n; j++) {
            temp = cal(i,j);
            ds.AddEdge(i,j,temp);
            ds.AddEdge(j,i,temp);
            dt.AddEdge(i,j,temp);
            dt.AddEdge(j,i,temp);
        }
    }
    ds.dijkstra(S);
    dt.dijkstra(T);
    double ansx = 0,ansy = 0;
    double ans = INF;
    int ax,ay;
    for(int i = S; i <= T; i++) {
        for(int j = S; j <= T; j++) {
            if(i == j) continue;
            if(i == S && j != T) temp = p[j].x;
            else if(i != S && j == T) temp = w - p[i].x;
            else if(i == S && j == T) temp = w;
            else temp = cal(i,j);
            if(ans > max(max(ds.dis[i],temp / 2),max(temp / 2,dt.dis[j]))) {
                ax = i;
                ay = j;
                ans = max(max(ds.dis[i],temp / 2),max(temp / 2,dt.dis[j]));
                if(i != S && j != T) {
                    ansx = (p[i].x + p[j].x) * 0.5;
                    ansy = (p[i].y + p[j].y) * 0.5;
                }
                if(i == S && j != T) {
                    ansx = p[j].x * 0.5;
                    ansy = p[j].y;
                }
                if(i != S && j == T) {
                    ansx = (p[i].x + w) * 0.5;
                    ansy = p[i].y;
                }
                if(i == S && j == T) {
                    ansx = w * 0.5;
                    ansy = 0;
                }
            }
        }
    }
    printf("%.12lf %.12lf\n",ansx,ansy);
    return 0;
}
/*
10 7
2 2
2 4
5 1
5 3
8 2
7 5
9 4
*/