HDU 4022 Bombing STL/二分

用stl的set和map

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<iterator>
using namespace std;
typedef map<int,multiset<int> > def;

void pop(def &a,def &b,int k)
{
	printf("%d\n",a[k].size());
	for(multiset<int>::iterator i=a[k].begin();i!=a[k].end();i++)
		b[*i].erase(k);
	a[k].clear();
}
int main()
{
	int n,m,x,y;
	while(1)
	{
		scanf("%d%d",&n,&m);
		if(n==0&&m==0)return 0;
		def h;def l;
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			h[x].insert(y);
			l[y].insert(x);
		}
		for(int j=0;j<m;j++)
		{
			scanf("%d%d",&x,&y);
		   if(x==0)
				pop(h,l,y);
			else
				pop(l,h,y);
		}
		printf("\n");
	}
	return 0;
}

或者二分

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100010;
struct Node
{
       int x,y;
}p[N];
bool hv[N];
int a[N],b[N];
bool cmp1(int k1,int k2)
{
     return p[k1].x < p[k2].x;
}
bool cmp2(int k1,int k2)
{
     return p[k1].y < p[k2].y;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) && !(n==0 && m==0))
    {
         memset(hv,true,sizeof(hv));
         for(int i=1;i<=n;i++) {
             scanf("%d%d",&p[i].x,&p[i].y);
             a[i]=b[i]=i;
         }
         sort(a+1,a+1+n,cmp1);
         sort(b+1,b+1+n,cmp2);
         while(m--)
         {
              int c,d;
              scanf("%d%d",&c,&d);
              int ans=0;
              if(c==0)
              {
                      int l=1,r=n,t;
                      while(l<r)
                      {
                           int mid=(l+r)/2;
                           if(p[a[mid]].x >= d) r=mid;
                           else l=mid+1;
                      }
                      t=l;
                      while(t<=n && p[a[t]].x == d) ans+=hv[a[t]],hv[a[t]]=false,t++;
              }
              else
              {
                      int l=1,r=n,t;
                      while(l<r)
                      {
                           int mid=(l+r)/2;
                           if(p[b[mid]].y >= d) r=mid;
                           else l=mid+1;
                      }
                      t=l;                      
                      while(t<=n && p[b[t]].y == d) ans+=hv[b[t]],hv[b[t]]=false,t++;
              }
              printf("%d\n",ans);
         }
         puts("");
    }
}