HDU 4606 occupy cities (calculation ry + Shortest Path coverage)

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = contents by --- cxlove

Question: N cities need to be occupied. M lines are obstacles and P soldiers can use them. Each soldier has a backpack, which can only be replenished once after the city is occupied. Ask the minimum size of the backpack. You can use the P soldiers to complete the task at any starting point.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4606

First, enumerate all vertices, find the distance, and determine whether it is consistent with the line segment. Then Floyd pre-process the Shortest Path

The second answer is used to determine whether the answer is reachable.

Edge construction is performed based on the order of occupation, and the result is determined based on the binary value.

Then, determine whether the minimum coverage is less than or equal to P.

#include <iostream>#include <queue>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>using namespace std;typedef long long LL;const int N = 505;const int M = 3000000;const int INF = 100000;const double eps = 1e-7;inline int dcmp(double d){    return d < -eps? -1 : d > eps;}inline double sqr(double d){    return d * d;}struct Point {    double x, y;    Point (){}    Point (double _x,double _y):x(_x),y(_y){}    inline bool operator == (const Point &p)const {        return (dcmp(x - p.x) == 0 && dcmp(y - p.y) == 0);    }    inline Point operator - (const Point &p)const {        return Point(x - p.x,y - p.y);    }    inline double operator * (const Point &p)const {        return x * p.y - y * p.x;    }    inline double operator / (const Point &p)const {        return x * p.x + y * p.y;    }    inline double Distance(Point p){        return sqrt(sqr(p.x - x) + sqr(p.y - y));    }    void input(){        scanf ("%lf %lf", &x, &y);    }}p[N];struct Line{    Point a,b;    void input(){        a.input();        b.input();    }    Line(){}    Line(Point _a,Point _b):a(_a),b(_b){}    inline bool operator == (const Line &l) const{        return (a == l.a && b == l.b) || (a == l.b && b == l.a);    }    inline double operator * (const Point &p)const {        return (b - a) * (p - a);    }    inline double operator / (const Point &p)const {        return (p - a) / (p - b);    }    inline int SegCrossSeg(const Line &v){        int d1 = dcmp((*this) * v.a);        int d2 = dcmp((*this) * v.b);        int d3 = dcmp(v * a);        int d4 = dcmp(v * b);        if((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;        return ((d1 == 0 && dcmp((*this) / v.a) <= 0)            ||(d2 == 0 && dcmp((*this) / v.b) <= 0)            ||(d3 == 0 && dcmp(v / a) <= 0)            ||(d4 == 0 && dcmp(v / b) <= 0) );    }}l[N];int n , m , K , seq[N];int match[N], vis[N];double dist[N][N];int tot , start[N];struct Edge {    int v, next;}e[N * N];void _add (int u , int v) {    e[tot].v = v;    e[tot].next = start[u];    start[u] = tot ++;}bool dfs (int u) {    for (int i = start[u] ; i != -1 ; i = e[i].next) {        int v = e[i].v;        if (! vis[v]) {            vis[v] = 1;            if(match[v] == -1 || dfs(match[v])) {                match[v] = u;                return true;            }        }    }    return false;}int hungry () {    int ans = 0;    memset (match , -1, sizeof(match)) ;    for (int i = 1 ; i <= n ; i ++) {        memset (vis , 0 , sizeof(vis));        if (dfs(i)) ans ++;    }    return ans;}bool check (double mid) {    tot = 0 ;    memset (start , -1 , sizeof(start));    for (int i =1 ; i <= n ; i ++) {        for (int j = i + 1 ; j <= n ; j ++) {            if(dist[seq[i]][seq[j]] <= mid)                 _add (seq[i] , seq[j] + n);        }    }    return n - hungry() <= K;}double solve(){    double l = 0 , r = INF;    double ans = -1;    int cnt = 50;    while(cnt --){        double mid = (l + r) * 0.5;        if(check(mid)){            ans = mid;            r = mid;        }        else l = mid;    }    return ans;}int main(){    int t;    scanf ("%d", &t);    while (t --){        scanf ("%d%d%d", &n, &m, &K);        for (int i = 1 ; i <= n ; i ++){            p[i].input();        }        for (int i = 1 ; i <= m ; i ++){            l[i].input();            p[n + (i - 1) * 2 + 1] = l[i].a;            p[n + i * 2] = l[i].b;        }        for (int i = 1 ; i <= n + m * 2 ; i ++){            for (int j = 1 ; j <= n + m * 2 ; j ++){                if(i == j) dist[i][j] = 0.0;                else {                    bool flag = false;                    for (int k = 1 ; k <= m ; k ++){                        if(l[k] == Line(p[i] , p[j])) continue;                        if (l[k].SegCrossSeg(Line(p[i] , p[j])) == 2) {                            flag = true;                            break ;                         }                    }                    if(flag) dist[i][j] = 1e9;                    else dist[i][j] = p[i].Distance(p[j]);                }            }        }        for(int i = 1;i <= n;i ++){            scanf("%d",&seq[i]);        }        for (int k = 1 ; k <= n + m * 2 ; k ++) {            for (int i = 1 ; i <= n + m * 2 ; i ++) {                for (int j = 1 ; j <= n + m * 2 ;j ++ ){                    dist[i][j] = fmin(dist[i][j] , dist[i][k] + dist[k][j]);                }            }        }        double ans = solve();        printf("%.2f\n",ans);    }    return 0;}