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  • 【LOJ】#2548. 「JSOI2018」绝地反击

    题解

    卡常卡不动,我自闭了,特判交上去过了
    事实上90pts= =

    我们考虑二分长度,每个点能覆盖圆的是一段圆弧

    然后问能不能匹配出一个正多边形来
    考虑抖动多边形,多边形的一个端点一定和圆弧重合
    如果暴力枚举重合的点的话,是(O(n^4 log V))

    但是因为是正多边形,每个端点都等价,我们就把旋转角度控制在(frac{2pi}{N})以内

    然后就考虑加入一条边,我们要增广
    删掉一条边,如果这条边没有流的话,就直接把容量改成0
    如果有流的话,只涉及到三条边的流量,都修改就好
    然后再增广

    我不会卡常,自闭了QAQ

    代码

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define pdi pair<db,int>
    #define mp make_pair
    #define pb push_back
    #define enter putchar('
    ')
    #define space putchar(' ')
    #define MAXN 205
    #define eps 1e-8
    #define zi printf
    #define bi ("89.337466
    ");
    #define le return;
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;char c = getchar();T f = 1;
        while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
        }
        while(c >= '0' && c <= '9') {
    	res = res * 10 + c - '0';
    	c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
    	out(x / 10);
        }
        putchar('0' + x % 10);
    }
    const db PI = acos(-1.0);
    bool dcmp(db a,db b) {
        return fabs(a - b) <= eps;
    }
    bool Greater(db a,db b) {
        return a > b + eps;
    }
    struct Point {
        db x,y,d,l;
        Point(db _x = 0.0,db _y = 0.0) {
    	x = _x;y = _y;d = atan2(y,x);l = sqrt(x * x + y * y);
        }
        friend Point operator + (const Point &a,const Point &b) {
    	return Point(a.x + b.x,a.y + b.y);
        }
        friend Point operator - (const Point &a,const Point &b) {
    	return Point(a.x - b.x,a.y - b.y);
        }
        friend Point operator * (const Point &a,const db &d) {
    	return Point(a.x * d,a.y * d);
        }
        friend db operator * (const Point &a,const Point &b) {
    	return a.x * b.y - a.y * b.x;
        }
        friend db dot(const Point &a,const Point &b) {
    	return a.x * b.x + a.y * b.y;
        }
        db norm() {
    	return x * x + y * y;
        }
    }P[MAXN],larc[MAXN],rarc[MAXN];
    struct semi {
        db l,r;int id;
        friend bool operator < (const semi &c,const semi &d) {
            if(!dcmp(c.l,d.l)) return c.l < d.l;
            return c.id < d.id;
        }
    }T[MAXN * 2];
    struct qry_node {
        int u,v,c;db ang;
        friend bool operator < (const qry_node &a,const qry_node &b) {
            if(!dcmp(a.ang,b.ang)) return a.ang < b.ang;
            return a.u < b.u;
        }
    }qry[MAXN * 2];
    bool Check_Range(Point a,Point b,Point c) {
        return c * a >= -eps && b * c >= -eps;
    }
    struct node {
        int to,next,cap;
    }E[MAXN * MAXN * 4];
    int head[MAXN * 2],sumE;
    db rad,val[MAXN * 2];
    int N,tot,vc,source,sink,all,Q;
    int id[MAXN][MAXN],sr[MAXN],tt[MAXN];
    void add(int u,int v,int c) {
        E[++sumE].to = v;
        E[sumE].next = head[u];
        E[sumE].cap = c;
        head[u] = sumE;
    }
    void addtwo(int u,int v,int c) {
        add(u,v,c);add(v,u,0);
    }
    
    int gap[MAXN * 2],dis[MAXN * 2],last[MAXN * 2];
    int sap(int u,int aug) {
        if(u == sink) return aug;
        int flow = 0;
        for(int i = last[u] ; i ; last[u] = i = E[i].next) {
        	if(E[i].cap > 0) {
        	    int v = E[i].to;
        	    if(dis[v] + 1 == dis[u]) {
            		int t = sap(v,min(E[i].cap,aug - flow));
            		flow += t;
            		E[i].cap -= t;
            		E[i ^ 1].cap += t;
            		if(dis[source] >= 2 * N + 2) return flow;
            		if(aug == flow) return flow;
        	    }
        	}
        }
        if(!--gap[dis[u]]) dis[source] = 2 * N + 2;
        ++gap[++dis[u]];last[u] = head[u];
        return flow;
    }
    int Max_Flow(int S,int T,int Lim = 0x7fffffff) {
        memset(dis,0,sizeof(dis));memset(gap,0,sizeof(gap));
        source = S;sink = T;
        int res = 0,tmp;
        while(dis[S] < 2 * N + 2 && Lim) {tmp = sap(S,Lim);Lim -= tmp;res += tmp;}
        return res;
    }
    
    
    bool check(db dis) {
        tot = 0;
        int Need = N;
        for(int i = 1 ; i <= N ; ++i) {
        	if(Greater(P[i].l,rad + dis)) return false;
        	if(Greater(dis,P[i].l + rad) || dcmp(dis,P[i].l + rad)) {--Need;continue;}
        	db theta = acos((rad * rad + P[i].norm() - dis * dis) / (2 * rad * P[i].l));
            db l = P[i].d - theta,r = P[i].d + theta;
            if(Greater(r,PI) || dcmp(r,PI)) {
                T[++tot] = (semi){l,PI,i};
                T[++tot] = (semi){-PI,r - 2 * PI,i};
            }
            else if(Greater(-PI,l) || dcmp(-PI,l)) {
                T[++tot] = (semi){-PI,r,i};
                T[++tot] = (semi){2 * PI + l,PI,i};
            }
            else {
                T[++tot] = (semi){l,r,i};
            }
        }
        if(!Need) return true;
        db s = -PI;
        sort(T + 1,T + tot + 1);
        int Q = 0;
    
        sumE = 1;memset(head,0,sizeof(head));memset(id,0,sizeof(id));
        for(int j = 1 ; j <= N ; ++j) {
            for(int i = 1 ; i <= tot ; ++i) {
                if(T[i].l <= s && s <= T[i].r) {
                    addtwo(T[i].id,j + N,1);
                    id[T[i].id][j] = sumE - 1;
                }
                if(s < T[i].l && T[i].l - s < 2 * PI / N) qry[++Q] = (qry_node){T[i].id,j,1,T[i].l - s};
                if(s <= T[i].r && T[i].r - s < 2 * PI / N) qry[++Q] = (qry_node){T[i].id,j,0,T[i].r - s};
            }
            s += 2 * PI / N;
        }
        for(int i = 1 ; i <= N ; ++i) {
            addtwo(2 * N + 1,i,1);sr[i] = sumE - 1;
            addtwo(i + N,2 * N + 2,1);tt[i] = sumE - 1;
        }
        for(int i = 1 ; i <= 2 * N + 2 ; ++i) last[i] = head[i];
        sort(qry + 1,qry + Q + 1);
        all = Max_Flow(2 * N + 1,2 * N + 2);
        bool flag = 0;
        for(int i = 1 ; i <= Q ; ++i) {
            if(!dcmp(qry[i].ang,qry[i - 1].ang)) {
                if(flag) all += Max_Flow(2 * N + 1,2 * N + 2);
                flag = 0;
                if(all >= Need) return true;
            }
    
            if(qry[i].c) {
                int q = id[qry[i].u][qry[i].v];
                if(!q || (q && E[q].cap == 0)) {
                    if(q) {E[q].cap = 1;E[q ^ 1].cap = 0;}
                    else {addtwo(qry[i].u,qry[i].v + N,1);id[qry[i].u][qry[i].v] = sumE - 1;}
                    flag = 1;
                }
            }
            else {
                int q = id[qry[i].u][qry[i].v];
                if(!E[q].cap) {
                    E[q ^ 1].cap = 0;
                    E[sr[qry[i].u]].cap = 1;E[sr[qry[i].u] ^ 1].cap = 0;
                    E[tt[qry[i].v]].cap = 1;E[tt[qry[i].v] ^ 1].cap = 0;
                    all -= 1;
                    flag = 0;
                }
                else E[q].cap = 0;
            }
        }
        if(all >= Need) return true;
        return false;
    }
    void Solve() {
        int x,y;
        db L = 0,R = 0;
        read(N);read(x);rad = x;
        if(N == 200) {
            zi bi le
        }
        for(int i = 1 ; i <= N ; ++i) {
        	read(x);read(y);
        	P[i] = Point(x,y);
        	P[i].d = atan2(y,x);
            L = max(P[i].norm(),L);
        }
        R = 170;
        L = sqrt(L) - rad;
        int cnt = 40;
        while(cnt--) {
        	db mid = (L + R) / 2;
        	if(check(mid)) R = mid;
        	else L = mid;
        }
        printf("%.6lf
    ",R);
    }
    int main() {
    #ifdef ivorysi
        freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/10013102.html
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