ZOJ-2362 Beloved Sons 贪心 | KM
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2362
裸的匹配问题，直接KM，就算是O(n^4)的KM也不会超。当然注意到题目中左边的点到右点所连的边的权值是一样的，所以完全可以贪心拍个序，然后找增广路。。。
1 //STATUS:C++_AC_250MS_848KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 //typedef __int64 LL; 33 //typedef unsigned __int64 ULL; 34 //const 35 const int N=410; 36 const int INF=0x3f3f3f3f; 37 const int MOD=100000,STA=8000010; 38 //const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 58 struct Node{ 59 int val,id; 60 bool operator < (const Node& a)const{ 61 return val>a.val; 62 } 63 }nod[N]; 64 int ca; 65 66 int w[N][N],y[N],vis[N]; 67 int n,m; 68 69 int dfs(int u) 70 { 71 int v; 72 for(v=1;v<=n;v++){ 73 if(w[u][v] && !vis[v]){ 74 vis[v]=1; 75 if(y[v]==-1 || dfs(y[v])){ 76 y[v]=u; 77 return 1; 78 } 79 } 80 } 81 return 0; 82 } 83 84 int main() 85 { 86 // freopen("in.txt","r",stdin); 87 int i,j,tot,a; 88 int x[N]; 89 scanf("%d",&ca); 90 while(ca--) 91 { 92 scanf("%d",&n); 93 for(i=1;i<=n;i++){ 94 scanf("%d",&nod[i].val); 95 nod[i].val*=nod[i].val; 96 nod[i].id=i; 97 } 98 mem(w,0); 99 for(i=1;i<=n;i++){ 100 scanf("%d",&tot); 101 while(tot--){ 102 scanf("%d",&a); 103 w[i][a]=nod[i].val; 104 } 105 } 106 sort(nod+1,nod+n+1); 107 108 mem(x,0); 109 mem(y,-1); 110 for(i=1;i<=n;i++){ 111 mem(vis,0); 112 dfs(nod[i].id); 113 } 114 for(i=1;i<=n;i++) 115 if(y[i]!=-1)x[y[i]]=i; 116 117 printf("%d",x[1]); 118 for(i=2;i<=n;i++) 119 printf(" %d",x[i]); 120 putchar('\n'); 121 } 122 return 0; 123 }
