可以看出蓝的之和一定比红的之和要大，也就是说符合条件的匹配一定是权值最小的，所以二分图的最佳完美匹配。。KM
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 110 12 #define LL long long 13 #define INF 9999999999.0 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct point 18 { 19 int x,y; 20 //point(int x=0,int y=0):x(x),y(y){} 21 }p[N<<1]; 22 double dis(point a,point b) 23 { 24 return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y)); 25 } 26 double dcmp(double x) 27 { 28 if(fabs(x)<eps) return 0; 29 return x<0?-1:1; 30 } 31 double w[N][N]; 32 int n,link[N],x[N],y[N]; 33 double lx[N],ly[N],slack[N],d; 34 bool match(int u) 35 { 36 x[u] = 1; 37 for(int i = 1; i <= n ; i++) 38 { 39 if(y[i]) continue; 40 double o = lx[u]+ly[i]-w[u][i]; 41 if(dcmp(o)==0) 42 { 43 y[i] = 1; 44 if(!link[i]||match(link[i])) 45 { 46 link[i] = u; 47 return true; 48 } 49 } 50 else if(slack[i]>o) slack[i] = o; 51 } 52 return false; 53 } 54 void km() 55 { 56 int i,j; 57 for(i = 1 ;i <= n ;i++) 58 lx[i] = -INF; 59 memset(ly,0,sizeof(ly)); 60 memset(link,0,sizeof(link)); 61 for(i = 1; i <= n ;i++) 62 for(j = 1; j <= n ;j++) 63 lx[i] = max(lx[i],w[i][j]); 64 for(i = 1; i <= n ;i++) 65 { 66 for(j = 1; j <= n;j++) 67 slack[j] = INF; 68 for(;;) 69 { 70 d = INF; 71 memset(x,0,sizeof(x)); 72 memset(y,0,sizeof(y)); 73 if(match(i)) break; 74 for(j =1 ; j <= n ;j++) 75 if(!y[j]&&d>slack[j]) d =min(d,slack[j]); 76 for(j =1 ;j <= n ;j++) if(x[j]){lx[j]-=d;} 77 for(j =1 ;j <= n ;j++) if(y[j]){ly[j]+=d;} 78 // else slack[j]-=d; 79 //cout<<","; 80 } 81 } 82 } 83 84 int main() 85 { 86 int i,j; 87 while(scanf("%d",&n)!=EOF) 88 { 89 //init(); 90 for(i = 1; i <=n+n; i++) 91 scanf("%d%d",&p[i].x,&p[i].y); 92 for(i =1; i <= n ;i++) 93 for(j = 1; j <= n ; j++) 94 { 95 w[i][j] = -dis(p[i+n],p[j]); 96 } 97 km(); 98 for(i = 1; i <= n ;i++) 99 printf("%d ",link[i]); 100 } 101 return 0; 102 }