poj 2482 挑战 284

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
struct Ne{
  int x,y,v;
}ne[10005];

int data[1<<28];

bool cmp(Ne a,Ne b)
{
    return a.y<b.y;
}

void build(int k,int l,int r)
{
    data[k]=0;
    if(r-l==1)
      return;
    build(2*k+1,l,(l+r)>>1);
    build(2*k+2,(l+r)>>1,r);
}

void update(int k)
{
     while(k>0)
     {
         k=(k-1)/2;
         data[k]=max(data[2*k+1],data[2*k+2]);
     }
}

void add(int k,int a,int b,int l,int r,int v)
{
     //cout<<"////3"<<endl;
     if(r<=a||l>=b)
        {
            //printf("1\n");
            return;
        }
     else if(l<=a&&b<=r)
     {
        data[k]+=v;
        update(k);
        //printf("2\n");
        return;
     }
     else if(r>a&&l<b)
     {
       // printf("4\n");
        add(2*k+1,a,(a+b)>>1,l,r,v);
        add(2*k+2,(a+b)>>1,b,l,r,v);
     }
}

int main()
{
    int n,a,b,rr;
    ne[0].y=0;
    while(~scanf("%d %d %d",&n,&a,&b))
    {
        rr=0;
        for(int i=1;i<=n;i++)
                {
                    scanf("%d %d %d",&ne[i].x,&ne[i].y,&ne[i].v);
                    rr=max(rr,ne[i].x);
                }
        sort(ne+1,ne+n+1,cmp);
        build(0,0,rr+1);
        int ans=0,l=0,r=0;
        for(;;)
        {
            //cout<<"///1"<<endl;
            while(ne[r].y-ne[l].y<b)
            {
                r++;
                if(r>n)
                    break;
                //cout<<"////2"<<endl;
                int k=ne[r].x-a+1;
                if(k<0)  k=0;
                //printf("%d %d\n",k,ne[r].x+a);
                add(0,0,rr+1,k,ne[r].x+a,ne[r].v);
                ans=max(data[0],ans);
                //cout<<"///ans:"<<ans<<endl;
            }
            if(r>n) break;
            add(0,1,rr+1,ne[l].x-a+1,ne[l].x+a,-ne[l].v);
            l++;
        }
        printf("%d\n",ans);
    }
    return 0;
}