POJ 1034 The Dog Task

由于在每个顶点只能走出去寻找一个好玩的地方，可以将顶点与每个符合条件的好玩的地方匹配，求最大匹配即可

 

一开始由于没看到限制条件一个，没想到用匹配做

 

 

代码：

#include<iostream>#include<cmath>using namespace std;#define MAX 105int master[MAX][2],dog[MAX][2],u[MAX],link[MAX];int g[MAX][MAX],res[MAX];int n,m;double getLen(int x1,int y1,int x2,int y2){double x=(double)(x1-x2),y=(double)(y1-y2);return sqrt(x*x+y*y);}int dfs(int t){for(int i=1;i<=m;i++){if(!u[i]&&g[t][i]){u[i]=1;if(link[i]==-1||dfs(link[i])){link[i]=t;res[t]=i;return 1;}}}return 0;}int hungary(){int i,num=0;memset(link,-1,sizeof(link));memset(res,-1,sizeof(res));for(i=1;i<=n;i++){memset(u,0,sizeof(u));num+=dfs(i);}return num;}int main(){int i,j;double len1,len2,len;scanf("%d %d",&n,&m);for(i=1;i<=n;i++){scanf("%d %d",&master[i][0],&master[i][1]);}for(j=1;j<=m;j++){scanf("%d %d",&dog[j][0],&dog[j][1]);}for(i=1;i<n;i++){for(j=1;j<=m;j++){len1=getLen(master[i][0],master[i][1],dog[j][0],dog[j][1]);len2=getLen(master[i+1][0],master[i+1][1],dog[j][0],dog[j][1]);len=getLen(master[i][0],master[i][1],master[i+1][0],master[i+1][1]);//cout<<len1<<" "<<len2<<" "<<len*2<<endl;if(len1+len2<=2*len)g[i][j]=1;}}printf("%d/n",n+hungary());for(i=1;i<=n;i++){printf("%d %d ",master[i][0],master[i][1]);if(res[i]!=-1){printf("%d %d ",dog[res[i]][0],dog[res[i]][1]);}}printf("/n");return 0;}