[zoj]第十九届浙大校赛
Thanks, TuSimple!
Description
As a manager of TuSimple, you are going to hold a dancing party for both the Development Department and the Marketing Department. There will be gentlemen and ladies in total and they are going to dance in pairs. After a careful investigation, we have already known that for each person, they like to dance with either a taller person or a person with smaller height. To simplify the problem, there are no two persons of the same height and people are only allowed to dance with a person of the opposite gender. In order to reserve a proper dancing field, you must calculate the maximum possible number of pairs of people dancing at the same time.
正确解法:
简单模拟然后居然没有写出来QAQ
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <map> 8 #include <vector> 9 using namespace std; 10 typedef long long ll; 11 const int inf=0x7fffffff; 12 const int N=100000+100; 13 int T,n,m,x; 14 ll a[N],b[N],nv0[N],nv1[N],nan1[N],nan0[N]; 15 int aa,bb,cc,dd; 16 int main() 17 { 18 scanf("%d",&T); 19 while(T--) 20 { 21 int ans=0; 22 memset(a,0,sizeof(a)); 23 memset(b,0,sizeof(b)); 24 aa=bb=cc=dd=0; 25 scanf("%d %d",&n,&m); 26 for(int i=1;i<=n;i++) 27 scanf("%lld",&a[i]); 28 for(int i=1;i<=m;i++) 29 scanf("%lld",&b[i]); 30 for(int i=1;i<=n;i++) 31 { 32 scanf("%d",&x); 33 if(x==1) nan1[++bb]=a[i]; 34 else nan0[++aa]=a[i]; 35 } 36 for(int i=1;i<=m;i++) 37 { 38 scanf("%d",&x); 39 if(x==1) nv1[++dd]=b[i]; 40 else nv0[++cc]=b[i]; 41 } 42 sort(nan0+1,nan0+aa+1); 43 sort(nan1+1,nan1+bb+1); 44 sort(nv0+1,nv0+cc+1); 45 sort(nv1+1,nv1+dd+1); 46 for(int i=1,j=1;i<=aa&&j<=dd;) 47 { 48 if(nan0[i]>nv1[j]) 49 { 50 ans++; i++; j++; 51 } 52 else i++; 53 } 54 for(int i=1,j=1;i<=bb&&j<=cc;) 55 { 56 if(nan1[i]<nv0[j]) 57 { 58 ans++; i++; j++; 59 } 60 else i++; 61 } 62 printf("%d\n",ans); 63 } 64 65 66 return 0; 67 }
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