uva 11992 - Fast Matrix Operations

简单的线段树的题；

有两种方法写这个题，目前用的熟是这种慢点的；

不过不知道怎么老是T；

感觉网上A过的人的时间度都好小，但他们都是用数组实现的

难道是指针比数组慢？

好吧，以后多用数组写写吧！

超时的代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define maxn 1000009
 using namespace std;

 struct node
 {
     int l,r;
     int ma,mi,sum;
     int ad,st;
     bool flag1,flag2;
     node *left,*right;
 } no[maxn*];
 int nonocount;

 void build(node *rt,int l,int r)
 {
     rt->l=l;
     rt->r=r;
     rt->sum=;
     rt->ma=;
     rt->mi=;
     if(l==r)
         return;
     int mid=(l+r)>>;
     rt->left=no+nonocount++;
     rt->right=no+nonocount++;
     build(rt->left,l,mid);
     build(rt->right,mid+,r);
 }

 void maintain(node *rt)
 {
     rt->ma=max(rt->left->ma,rt->right->ma);
     rt->mi=min(rt->left->mi,rt->right->mi);
     rt->sum=rt->left->sum+rt->right->sum;
 }

 void pushdown(node *rt)
 {
     if(rt->flag1==)
     {
         rt->flag1=;
         rt->ma+=rt->ad;
         rt->mi+=rt->ad;
         rt->sum+=(rt->r-rt->l+)*rt->ad;
         if(rt->l!=rt->r)
         {
             rt->left->ad=rt->right->ad=rt->ad;
             rt->left->flag1=rt->right->flag1=;
             pushdown(rt->left);
             pushdown(rt->right);
         }
         rt->ad=;
         return;
     }
     if(rt->flag2==)
     {
         rt->flag2=;
         rt->sum=(rt->r-rt->l+)*rt->ad;
         rt->ma=rt->mi=rt->ad;
         if(rt->l!=rt->r)
         {
             rt->left->ad=rt->right->ad=rt->ad;
             rt->left->flag2=rt->right->flag2=;
             pushdown(rt->left);
             pushdown(rt->right);
         }
         rt->ad=;
         return;
     }
 }

 void add(node *rt,int l,int r,int v)
 {
     if(l==rt->l&&r==rt->r)
     {
         rt->ad=v;
         rt->flag1=;
         pushdown(rt);
         return;
     }
     int mid=(rt->l+rt->r)>>;
     if(r<=mid)add(rt->left,l,r,v);
     else if(l>=mid+)add(rt->right,l,r,v);
     else
     {
         add(rt->left,l,mid,v);
         add(rt->right,mid+,r,v);
     }
     maintain(rt);
 }

 void set(node *rt,int l,int r,int v)
 {
     if(l==rt->l&&r==rt->r)
     {
         rt->ad=v;
         rt->flag2=;
         pushdown(rt);
         return;
     }
     int mid=(rt->l+rt->r)>>;
     if(r<=mid)set(rt->left,l,r,v);
     else if(l>=mid+)set(rt->right,l,r,v);
     else
     {
         set(rt->left,l,mid,v);
         set(rt->right,mid+,r,v);
     }
     maintain(rt);
 }
 int ssum,mma,mmi;
 void query(node *rt,int l,int r)
 {
     if(l==rt->l&&r==rt->r)
     {
         ssum+=rt->sum;
         mma=max(rt->ma,mma);
         mmi=min(rt->mi,mmi);
         return;
     }
     int mid=(rt->l+rt->r)>>;
     if(r<=mid)query(rt->left,l,r);
     else if(l>=mid+)query(rt->right,l,r);
     else
     {
         query(rt->left,l,mid);
         query(rt->right,mid+,r);
     }
 }

 int main()
 {
     int n,m,q,cmd,x1,y1,x2,y2,v;
     while(scanf("%d%d%d",&n,&m,&q)!=EOF)
     {
         nonocount=;
         for(int i=; i<=n; i++)
             build(no+i,,m);
         while(q--)
         {
             scanf("%d",&cmd);
             if(cmd==)
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 scanf("%d",&v);
                 for(int i=x1; i<=x2; i++)
                     add(no+i,y1,y2,v);
             }
             else if(cmd==)
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 scanf("%d",&v);
                 for(int i=x1; i<=x2; i++)
                     set(no+i,y1,y2,v);
             }
             else
             {
                 int ans=;
                 mma=-,mmi=;
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 for(int i=x1; i<=x2; i++)
                 {
                     ssum=;
                     query(no+i,y1,y2);
                     ans+=ssum;
                 }
                 printf("%d %d %d\n",ans,mmi,mma);
             }
         }
     }
     return ;
 }