题目链接：
http://acm.hust.edu.cn/vjudge/problem/307216
Froggy Ford
题意
青蛙过河，河中有若干个石头，现在你可以加一个石头，使得青蛙从左岸跳到右岸的最大跳跃距离最小。
题解
把左岸和右岸作为两个虚节点，用kruskal的思路处理出每个点到左岸需要跳跃的最大距离st[i]（最优情况下）和每个点到右岸的最大距离ed[i]，然后枚举两个点，在这两个点的中点放一个石头，得到ma=max(st[i],ed[j],dis(i,j)/2)，如果这个值比答案更优，就更新答案，同时记录新放的石头的坐标。
代码
#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;const int INF=0x3f3f3f3f;
const LL INFL=9e18;
const double eps=1e-8;const int maxn=1111;LL w,n;pair<LL,LL> pt[maxn];LL dis(LL x1,LL y1,LL x2,LL y2){return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}struct Edge{int u,v; LL w;Edge(int u,int v,LL w):u(u),v(v),w(w){}Edge(){}bool operator < (const Edge& tmp) const {return w<tmp.w;}
}egs[maxn*maxn];int tot;
int fa[maxn];
int find(int x){return fa[x]=fa[x]==x?x:find(fa[x]);
}vector<int> G[maxn];
int st[maxn],ed[maxn];void init(){clr(st,-1);clr(ed,-1);tot=0;rep(i,0,maxn) fa[i]=i;rep(i,0,maxn) G[i].push_back(i);
}int main() {freopen("froggy.in","r",stdin);freopen("froggy.out","w",stdout);init();scanf("%I64d%I64d",&w,&n);rep(i,1,n+1) scanf("%I64d%I64d",&pt[i].X,&pt[i].Y);rep(i,1,n+1){egs[tot++]=Edge(0,i,pt[i].X*pt[i].X);}rep(i,1,n+1){egs[tot++]=Edge(i,n+1,(w-pt[i].X)*(w-pt[i].X)); }rep(i,1,n+1){rep(j,i+1,n+1){egs[tot++]=Edge(i,j,dis(pt[i].X,pt[i].Y,pt[j].X,pt[j].Y));}}egs[tot++]=Edge(0,n+1,w*w);egs[tot++]=Edge(0,0,0);egs[tot++]=Edge(n+1,n+1,0);sort(egs,egs+tot);rep(i,0,tot){int u=egs[i].u,v=egs[i].v;if(u==0&&v==0){st[0]=i; continue;}if(u==n+1&&v==n+1){ed[n+1]=i; continue;}int pu=find(u);int pv=find(v);if(pu!=pv){if(find(0)==pu){rep(j,0,G[pv].size()){int tt=G[pv][j];st[tt]=i;}}else if(find(0)==pv){rep(j,0,G[pu].size()){int tt=G[pu][j];st[tt]=i;}}if(find(n+1)==pu){rep(j,0,G[pv].size()){int tt=G[pv][j];ed[tt]=i;}}else if(find(n+1)==pv){rep(j,0,G[pu].size()){int tt=G[pu][j];ed[tt]=i;}}fa[pv]=pu;while(G[pv].size()>0){G[pu].push_back(G[pv][G[pv].sz()-1]);G[pv].pop_back();}}}int ans_i=0,ans_j=0;double mi=9000000000000000000;rep(i,0,n+2){rep(j,0,n+2){if(i==j) continue;double tmp=max(egs[st[i]].w,egs[ed[j]].w);int ti=i,tj=j;if(ti>tj) swap(ti,tj);if(ti==0&&tj==n+1){tmp=max(tmp,w*w*1.0/4);}else if(ti==0){tmp=max(tmp,pt[tj].X*pt[tj].X*1.0/4);}else if(tj==n+1){tmp=max(tmp,(w-pt[ti].X)*(w-pt[ti].X)*1.0/4);}else{tmp=max(tmp,dis(pt[i].X,pt[i].Y,pt[j].X,pt[j].Y)*1.0/4);}if(tmp<mi){mi=tmp;ans_i=i; ans_j=j;}}}double ans_x=0,ans_y=0;if(ans_i>ans_j) swap(ans_i,ans_j);if(ans_i==0&&ans_j==n+1){ans_x=w*1.0/2;ans_y=0;}else if(ans_i==0){ans_x=pt[ans_j].X*1.0/2;ans_y=pt[ans_j].Y;}else if(ans_j==n+1){ans_x=(w+pt[ans_i].X)*1.0/2;ans_y=pt[ans_i].Y;}else{ans_x=(pt[ans_i].X+pt[ans_j].X)*1.0/2;ans_y=(pt[ans_i].Y+pt[ans_j].Y)*1.0/2; }printf("%.3lf %.3lf\n",ans_x,ans_y);return 0;
}
)
