[SDOI2012]拯救小云公主【bfs+二分答案】

题目链接

  正难则反。

  要直接求从起点到终点的最大距离，不妨反过来求最小的可以阻止骑士从起点到终点的对于全体圆的最小半径。

  那么，就是阻止从左上角到右下角的所有相交圆，于是，就是要变成没有从左上角到右下角的相交圆才可以，那么不妨跑一个bfs来判断，我们二分答案半径，然后看，是否左边界和上边界的相交圆可以抵达下边界和右边界。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-4
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 3e3 + 7;
int N;
double row, line, ans;
struct node
{
    double x, y;
    node(double a=0, double b=0):x(a), y(b) {}
    inline void In() { scanf("%lf%lf", &x, &y); }
} a[maxN];
bool vis[maxN];
queue<int> Q;
inline double _Dis(node e1, node e2) { return sqrt((e1.x - e2.x) * (e1.x - e2.x) + (e1.y - e2.y) * (e1.y - e2.y)); }
inline bool bfs(double r)
{
    while(!Q.empty()) Q.pop();
    for(int i=1; i<=N; i++) vis[i] = false;
    for(int i=1; i<=N; i++)
    {
        if(_Dis(a[i], node(1, 1)) < r || _Dis(a[i], node(row, line)) < r) return false;
        if(a[i].x < 1. + r || a[i].y + r > line)
        {
            vis[i] = true;
            Q.push(i);
        }
    }
    int u;
    while(!Q.empty())
    {
        u = Q.front(); Q.pop();
        if(a[u].x + r > row || a[u].y < 1. + r) return false;
        for(int i=1; i<=N; i++)
        {
            if(vis[i]) continue;
            if(_Dis(a[i], a[u]) < 2. * r)
            {
                vis[i] = true;
                Q.push(i);
            }
        }
    }
    return true;
}
int main()
{
    scanf("%d%lf%lf", &N, &row, &line);
    for(int i=1; i<=N; i++) a[i].In();
    double L = 0., R = min(row, line), mid = 0.; ans = 0.;
    while(R - L >= eps)
    {
        mid = (L + R) / 2.;
        if(bfs(mid))
        {
            L = mid;
            ans = mid;
        }
        else R = mid;
    }
    printf("%.2lf\n", ans);
    return 0;
}