【HDOJ6638】Snowy Smile（线段树）

题意：一个二维平面上有n个点，每个点的坐标是（x[i]，y[i]），权值是w[i]

求一个矩形使得其中所有点的权值和最大，输出权值和

n<=2e3，x[i]，y[i]，w[i]的绝对值<=1e9

思路：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 #define N  110000
 #define M  4100000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;

 ll t1[N<<],t2[N<<],t3[N<<],t4[N<<];
 int x[N],y[N],w[N],X[N],Y[N];
 VII V[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void clear(int l,int r,int p)
 {
     t1[p]=t2[p]=t3[p]=t4[p]=;
     if(l==r) return;
     int mid=(l+r)>>;
     clear(l,mid,ls);
     clear(mid+,r,rs);
 }

 void pushup(int p)
 {
     t1[p]=t1[ls]+t1[rs];
     t2[p]=max(t2[ls],t1[ls]+t2[rs]);
     t3[p]=max(t3[rs],t1[rs]+t3[ls]);
     t4[p]=max(max(t4[ls],t4[rs]),t3[ls]+t2[rs]);
 }

 void update(int l,int r,int x,int v,int p)
 {
     if(l==r)
     {
         t1[p]+=v;
         t2[p]=t3[p]=t4[p]=max(t1[p],0ll);
         return;
     }
     int mid=(l+r)>>;
     if(x<=mid) update(l,mid,x,v,ls);
      else update(mid+,r,x,v,rs);
     pushup(p);
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     int cas=read();
     while(cas--)
     {
         int n=read();
         X[]=Y[]=;
         rep(i,,n)
         {
             x[i]=read(),y[i]=read(),w[i]=read();
             X[++X[]]=x[i];
             Y[++Y[]]=y[i];
         }
         sort(X+,X+X[]+);
         sort(Y+,Y+Y[]+);
         X[]=unique(X+,X+X[]+)-X-;
         Y[]=unique(Y+,Y+Y[]+)-Y-;
         rep(i,,n)
         {
             x[i]=lower_bound(X+,X+X[]+,x[i])-X;
             y[i]=lower_bound(Y+,Y+Y[]+,y[i])-Y;
         }
         rep(i,,X[]) VII().swap(V[i]);
         rep(i,,n) V[x[i]].push_back(MP(y[i],w[i]));
         //printf("isok\n");
         ll ans=;
         rep(i,,X[])
         {
             clear(,Y[],);
             rep(j,i,X[])
             {
                 for(int k=;k<V[j].size();k++) update(,Y[],V[j][k].fi,V[j][k].se,);
                 ans=max(ans,t4[]);
             }
         }
         printf("%I64d\n",ans);
     }

     return ;
 }