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 }