UVA10080 Gopher II

题意：有n只地鼠(gopher)和m个地洞，现在遇到危险，地鼠要在s秒内钻进地洞里，地鼠的速度是v，但是每个地洞只能容纳1只地鼠。给定每只地鼠和每个地洞的位置，和s,v，求将会有危险的地鼠的个数。

分析：将地鼠看成一个点，所有地鼠看成一个集合X，将地洞看成一个点，所有地洞看成另一个集合Y，如果地鼠能跑到地洞就连一条线，变成二部图，题意就变成了求二部图最大匹配问题了，套用匈牙利算法。

代码：

//uva10080 Gopher II//AC By Warteac//2013-4-19//Runtime:0.036s#include<iostream>#include<cmath>#include<vector>#include<cstring>using namespace std;struct node{float x;float y;node(float a,float b){x = a; y = b;}float distance(node n){return sqrt((y-n.y)*(y-n.y)+(x-n.x)*(x-n.x));}};class GopherII{private:bool map[105][105];//邻接矩阵int nx,ny;//x,y点集的个数bool vis[105];//记录点有无被访问过    int link[105];//记录y集合到x集合映射关系    int vulgophers;    int dfs(int u);    int hungary();public:void initial();void readCase(int);void computing();void outResult();};int GopherII::dfs(int u){     for(int i=0;i<ny;i++){             if(map[u][i]&&!vis[i]){                  vis[i]=1;                  if(link[i]==-1||dfs(link[i])){//y值被访问过就到它对应的x值继续向下找增广路                        link[i]=u;//在递归返回过程中不断修改                        return 1;                  }             }     }     return 0;}int GopherII::hungary(){//匈牙利算法     int ans=0;     memset(link,-1,sizeof link);     for(int i=0;i<nx;i++){          memset(vis,0,sizeof vis);          ans+=dfs(i);     }     return ans;}void GopherII::initial(){memset(map,0,sizeof(map));memset(vis,0,sizeof(vis));memset(link,0,sizeof(link));vulgophers = 0;}void GopherII::readCase(int n){const double esp = 1e-10;int s,v;float x,y,x2,y2;vector <node> v1,v2;v1.clear();v2.clear();nx = n;cin >> ny >> s >> v;//cout << "s*v = "<<s*v <<endl;for(int i = 0 ; i < nx ; i++){cin >> x >> y;//cout <<"x = "<< x <<"y = "<< y <<endl;v1.push_back(node(x,y));}for(int i = 0 ; i < ny ; i++){cin >> x2 >> y2;//cout <<"x2 = "<< x2 <<"y2 = "<< y2 <<endl;v2.push_back(node(x2,y2));}//cout << "v1.size()= "<<v1.size()<<endl;//cout << "v2.size()= "<<v2.size()<<endl;for(int i = 0; i < v1.size(); i++){for(int j = 0; j < v2.size(); j++){    //cout << "v1[i].distance(v2[j]) = " << v1[i].distance(v2[j])<<endl;if(v1[i].distance(v2[j])+esp <= s*v){map[i][j] = 1;//map[j][i] = 1;}}}}void GopherII::computing(){vulgophers = nx - hungary();}void GopherII::outResult(){cout << vulgophers << endl;}int main(){GopherII g;int n;while(cin >> n){g.initial();g.readCase(n);g.computing();g.outResult();}}