题目大意

给定矩形中的\(n\)个点，矩形中的任意一个位置被离它最近的点控制，给出起点，求走出这个矩形最少被几个点控制过。
\(n \leq 600\)

解析

借用网上这张酷炫的图

可以发现，两个点\(i,j\)的垂直平分线把平面分为两部分，一部分归\(i\)控制，一部分归\(j\)控制，而垂直平分线上的点同时被\(i,j\)控制。

可见，只有经过两区域边界时答案才会增大。

那么把\(i\)与其它所有点的垂直平分线做一遍半平面交，得到\(i\)的控制区域，同时也能得到控制区域与\(i\)相邻的点。

于是我们构建一个图模型，对于控制区域相邻的\(i,j\)，连一条长度为\(1\)的边，对于靠着矩形边界的区域，向\(n+1\)连一条长度为\(1\)的边。输入数据保证一开始只被一个点控制，这个点一定离它最近，暴力求出来即可，然后跑最短路即可求出答案。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef double db;
const int N = 1007;
const db eps = 1e-6;

int tot, st[N], to[N * N * 2], nx[N * N * 2];
void add(int u, int v) { to[++tot] = v, nx[tot] = st[u], st[u] = tot; }
int S, T, head, tail, que[N], dis[N];
int spfa()
{
    memset(dis, 0, sizeof(dis));
    head = 1, que[tail = 1] = S, dis[S] = 1;
    while (head <= tail)
    {
        int u = que[head++];
        for (int i = st[u]; i; i = nx[i]) if (!dis[to[i]]) dis[to[i]] = dis[u] + 1, que[++tail] = to[i];
    }
    return dis[T] - 1;
}

int dcmp(db a, db b) { return fabs(a - b) < eps; }
struct VECTOR
{
    db x, y;
    VECTOR(db X = 0, db Y = 0) { x = X, y = Y; }
    db getang() { return atan2(y, x); }
};
db cross(VECTOR a, VECTOR b) { return a.x * b.y - a.y * b.x; }
VECTOR operator+(VECTOR a, VECTOR b) { return VECTOR(a.x + b.x, a.y + b.y); }
VECTOR operator-(VECTOR a, VECTOR b) { return VECTOR(a.x - b.x, a.y - b.y); }
VECTOR operator*(VECTOR a, db b) { return VECTOR(a.x * b, a.y * b); }
VECTOR operator/(VECTOR a, db b) { return VECTOR(a.x / b, a.y / b); }
typedef VECTOR POINT;
struct LINE
{
    POINT s, t;
    db ang;
    int tag;
    LINE() {}
    LINE(POINT S, POINT T, int ta) { s = S, t = T, ang = (t - s).getang(), tag = ta; }
};
int operator<(LINE a, LINE b) { return dcmp(a.ang, b.ang) ? cross(b.t - a.s, a.t - a.s) + eps > 0 : a.ang < b.ang; }
POINT inter(LINE a, LINE b) { return a.s + (a.t - a.s) * (cross(b.t - b.s, a.s - b.s) / cross(a.t - a.s, b.t - b.s)); }
int onright(POINT a, LINE b) { return cross(b.t - b.s, a - b.s) < 0; }
db sqr(db a) { return a * a; }
db getdis(db x1, db y1, db x2, db y2) { return sqrt(sqr(x1 - x2) + sqr(y1 - y2)); }

LINE l[N];
int total, n, X0, Y0, X1, Y1, x[N], y[N];
int cnt;

int h, t;
LINE q1[N];
POINT q2[N];

int main()
{
    scanf("%d", &total);
    while (total--)
    {
        memset(st, 0, sizeof(st));
        memset(nx, 0, sizeof(nx));
        scanf("%d%d%d%d%d", &n, &X1, &Y1, &X0, &Y0);
        if (!n) { printf("0\n"); continue; }
        for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
        for (int i = 1; i <= n; i++)
        {
            cnt = 0;
            l[++cnt] = LINE(POINT(0, 0), POINT(X1, 0), n + 1);
            l[++cnt] = LINE(POINT(X1, 0), POINT(X1, Y1), n + 1);
            l[++cnt] = LINE(POINT(X1, Y1), POINT(0, Y1), n + 1);
            l[++cnt] = LINE(POINT(0, Y1), POINT(0, 0), n + 1);
            for (int j = 1; j <= n; j++) if (j != i)
            {
                db mx = 1.0 * (x[i] + x[j]) / 2, my = 1.0 * (y[i] + y[j]) / 2;
                db tx = mx + my - y[j], ty = my + x[j] - mx;
                l[++cnt] = LINE(POINT(mx, my), POINT(tx, ty), j);
            }
            sort(l + 1, l + cnt + 1);
            h = 1, q1[t = 1] = l[1];
            for (int i = 2; i <= cnt; i++)
            {
                if (dcmp(l[i].ang, l[i - 1].ang)) continue;
                while (h < t && onright(q2[t - 1], l[i])) t--;
                while (h < t && onright(q2[h], l[i])) h++;
                q1[++t] = l[i];
                if (h < t) q2[t - 1] = inter(q1[t], q1[t - 1]);
            }
            while (h < t && onright(q2[t - 1], q1[h])) t--;
            while (h < t && onright(q2[h], q1[t])) h++;
            if (t - h <= 1) continue;
            for (int j = h; j <= t; j++) add(i, q1[j].tag);
        }
        S = 1;
        for (int i = 2; i <= n; i++) if (getdis(x[i], y[i], X0, Y0) < getdis(x[S], y[S], X0, Y0)) S = i;
        T = n + 1;
        printf("%d\n", spfa());
    }
    return 0;
}

标签：逃考,d%,db,st,int,SDOI2013,JZOJ3297,LINE,include
来源： https://www.cnblogs.com/zjlcnblogs/p/11116838.html