Red Huang

Red Huang

uva 11678

I personally think it's an easy problem.

Find two sequences that meet the rules.

Taking the minimum value is the answer.

//  
//        GGGGGGGGGGGGG        CCCCCCCCCCCCC               AAA  
//     GGG::::::::::::G     CCC::::::::::::C              A:::A  
//   GG:::::::::::::::G   CC:::::::::::::::C             A:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C            A:::::::A  
// G:::::G       GGGGGG C:::::C       CCCCCC           A:::::::::A  
//G:::::G              C:::::C                        A:::::A:::::A  
//G:::::G              C:::::C                       A:::::A A:::::A  
//G:::::G    GGGGGGGGGGC:::::C                      A:::::A   A:::::A  
//G:::::G    G::::::::GC:::::C                     A:::::A     A:::::A  
//G:::::G    GGGGG::::GC:::::C                    A:::::AAAAAAAAA:::::A  
//G:::::G        G::::GC:::::C                   A:::::::::::::::::::::A  
// G:::::G       G::::G C:::::C       CCCCCC    A:::::AAAAAAAAAAAAA:::::A  
//  G:::::GGGGGGGG::::G  C:::::CCCCCCCC::::C   A:::::A             A:::::A  
//   GG:::::::::::::::G   CC:::::::::::::::C  A:::::A               A:::::A  
//     GGG::::::GGG:::G     CCC::::::::::::C A:::::A                 A:::::A  
//        GGGGGG   GGGG        CCCCCCCCCCCCCAAAAAAA                   AAAAAAA  
#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <climits>  
#include <vector>  
#include <set>  
#include <map>  
#include <queue>  
#include <cctype>  
#include <utility>  
#include <ctime>  
using namespace std;  
#ifdef DEBUG  
#define VAR(a,b) decltype(b) a=(b)  
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)  
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);  
#else  
#define VAR(a,b) \_\_typeof(b) a=(b)  
#define debug(...)  
#define gettime()  
#endif  
typedef unsigned int uint;  
typedef long long int Int;  
  
#define Set(a,s) memset(a,s,sizeof(a))  
#define Write(w) freopen(w,"w",stdout)  
#define Read(r) freopen(r,"r",stdin)  
#define Pln() printf("\\n")  
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()  
#define De(x)printf(#x"%d\\n",x)  
#define For(i,x)for(int i=0;i<x;i++)  
#define CON(x,y) x##y  
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\  
    printf("%d ",dp\[htx\]\[hty\]);}Pln();}  
#define M 10005  
#define PII pair<int,int\>  
#define PB push\_back  
#define oo INT\_MAX  
#define Set\_oo 0x3f  
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)  
#define eps 1e-6  
clock\_t start\_time=clock(), end\_time;  
bool xdy(double x,double y){return x>y+eps;}  
bool xddy(double x,double y){return x>y-eps;}  
bool xcy(double x,double y){return x<y-eps;}  
bool xcdy(double x,double y){return x<y+eps;}  
int min3(int x,int y,int z){  
    int tmp=min(x,y);  
    return min(tmp,z);  
}  
int max3(int x,int y,int z){  
    int tmp=max(x,y);  
    return max(tmp,z);  
}  
int anum,bnum;  
bool ah\[100005\],bh\[100005\];  
int a\[M\],b\[M\];  
int used\[100005\];  
void solve(){  
    Set(used,0);  
    int aans=0,bans=0;  
    for(int i=0;i<anum;i++){  
        if(!bh\[a\[i\]\]){  
//            debug(">> %d\\n",a\[i\]);  
            bh\[a\[i\]\]=1;  
            aans++;  
        }  
    }  
    for(int i=0;i<bnum;i++){  
        if(!ah\[b\[i\]\]){  
//            debug("%d\\n",b\[i\]);  
            ah\[b\[i\]\]=1;  
            bans++;  
        }  
    }  
    printf("%d\\n",min(aans,bans));  
}  
int main() {  
    ios\_base::sync\_with\_stdio(0);  
    while(~scanf("%d%d",&anum,&bnum)&&anum+bnum){  
        Set(ah,0);  
        Set(bh,0);  
        for(int i=0;i<anum;i++){  
            scanf("%d",&a\[i\]);  
            ah\[a\[i\]\]=1;  
        }  
        for(int i=0;i<bnum;i++){  
            scanf("%d",&b\[i\]);  
            bh\[b\[i\]\]=1;  
        }  
        solve();  
    }  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
}  

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