Gym - 100851F Froggy Ford (Dijkstra)

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
*/

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