HDU 3016 Man Down

该题是一道，成段更新,单点查询(简单线段树+简单DP)，这个题与http://poj.org/problem?id=1436思想差不多，这里我就不重复了，这里就是要处理最大值的问题也就是DP；

我有的方法就是以该条边的两个端点往下搜索，如果下面有可见的边，选择值的那条边，然后再进行更新；

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
class Node
{
public:
    int l ,r , sum,cover;    
};
Node tree[2000024];
class segment
{
public:
    int x1,x2,y,value;    
};
segment seg[150024];
bool cmp( segment a, segment b )
{
   return a.y < b.y;
}
class Tree
{
public:
     void Maketree( int l , int r, int cnt );
     void Update( int l , int r ,int cnt, int id, int sum );
     int Max( int a, int b )
     { return a>b?a:b; }
     int Query( int cnt,int num  );
};

void Tree::Maketree( int l , int r , int cnt )
{
   tree[cnt].l = l ;
   tree[cnt].r = r;
   tree[cnt].cover = -1;
   tree[cnt].sum = 0;
   if( l == r ) 
        return ;
   int mid = ( l + r )>>1;
   Maketree( l , mid , cnt*2 );
   Maketree( mid + 1 , r , cnt*2+1 );    
}
void Tree::Update( int l , int r , int cnt , int id ,int sum)
{
   if( l<=tree[cnt].l&&r>=tree[cnt].r  )
   {
      tree[cnt].sum=sum+seg[id].value;
      tree[cnt].cover = id;
//      printf( "l==%d r==%d sum==%d\n",tree[cnt].l ,tree[cnt].r,tree[cnt].sum);
      return ;        
   }
   else
   {
        if( tree[cnt].cover !=-1  )//向下进行更新 
        {
           tree[cnt*2].cover = tree[cnt*2+1].cover = tree[cnt].cover;
           tree[cnt].cover = -1;
           tree[cnt*2].sum = tree[cnt*2+1].sum = tree[cnt].sum;    
        }
        int mid = ( tree[cnt].r + tree[cnt].l )>>1;
       if( l > mid ) Update( l ,r ,cnt*2+1, id ,sum );
       else
       {
         if( r <= mid ) Update( l ,r ,cnt * 2 , id ,sum );
         else
         {
              Update( l ,mid , cnt*2 , id ,sum );
              Update( mid + 1 , r , cnt*2+1 , id ,sum );        
         }
       }        
   }    
}
int Tree::Query( int cnt , int num )
{
    if( tree[cnt].cover!=-1 )
    {
//        printf( "%d %d\n",tree[cnt].sum,num );
        return tree[cnt].sum;    
    }    
    int mid = ( tree[cnt].l + tree[cnt].r  )>>1;
    if( num>mid ) return Query( cnt*2+1 , num );
    else return Query( cnt*2 , num );
    return 0;
}
int main( )
{
    int n;
    while( scanf( "%d",&n )==1 )
    {
       Tree e;
       e.Maketree( 0,200000 , 1  );
       for( int i = 0; i < n ; i++ )
       {

          scanf( "%d%d%d%d",&seg[i].y,&seg[i].x1,&seg[i].x2,&seg[i].value );
          seg[i].x1 *= 2;
          seg[i].x2 *= 2;        
       }
       sort( seg , seg+n , cmp ); //对Y进行排序 
       for( int i = 0 ; i < n-1 ; i++ )
       {
          int sum = e.Max( e.Query( 1 , seg[i].x1 ),e.Query( 1 , seg[i].x2 ) );//对两个端点进行选择最大值 
          e.Update( seg[i].x1 , seg[i].x2, 1 , i , sum );//进行更新 
         
       }
       int sum = 100 + e.Max( e.Query( 1 , seg[n-1].x1 ),e.Query( 1 , seg[n-1].x2 ) ) +seg[n-1].value;//最后一个段进行处理 
       printf( sum>0?"%d\n":"-1\n",sum);
    }
    return 0;    
}