Under Attack

Time Limit:  10 Seconds      Memory Limit:  65536 KB 

Doctor serves at a military air force base. One day, the enemy launch a sudden attack and the base is under heavy fire. The fighters in the airport must take off to intercept enemy bombers. However, the enemies know this clearly and they now focus on destroying
the runway. The situation is becoming worse rapidly! 

Every part of the runway has a damage level. On each bombing run, the damage level is increased by the bomb's damage . Fortunately, the enemy bombers has to stop bombing the runway when they run out of ammo. Then the ground crew have time to evaluate the situation
of runway so that they can come to repair the runway ASAP after enemy attacks. The most heavily-damaged part on fighters' taking off and landing path should first be repaired. Assume that runway start from north and head to south , and fighters will take off
or land only from north to south or vice versa. 

Now that the task is clear, the ground crew need the cooridinates of two points: first that is the most damaged point from north to south, second is the most damaged point from south to north.The base's central mainframe is down under hacker attack. So Doctor
could only use his poor little and shabby notebook to fulfill this task. Can you help him? 

Input

The input consists of multiple cases. 

The first line is the runway length L. L can be up to 15000.

 Next lines will describe enemy bombing runs ,each line describes effect range start end of each bombing run and enemy bomb damage d.if start is -1, this case ends..

There can be up to 3000 bombing run, each time the damage is up to 100. 

 Notice that the bombing range is from north to south, and runway range is [0,len].

Output

Output the cooridinates of two points: first that is the most damaged point from north to south, second is the most damaged point from south to north. 

Sample Input

10

1 5 2

6 9 2

-1 -1 -1

Sample Output

1 9

**

从早上起 这都一上午了，这个大水题最终a了！**

**

在不知道 最大覆盖次数求法之前，我先求出全长线段中最大值，也就是普通的区间更新加延迟标记。然后利用calculate函数从两边分别開始遍历找到左右最大值的位置。从多组測试数据上来看 。并没有什么差错，但就是wrong，后学会最大覆盖次数。1a。

见到的请帮我看看究竟是哪些 数据错了！！！！

**

**。！

**

第二段代码是正确地。

wrong code:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

int create(int root,int left,int right)
{
tree[root].left=left;
tree[root].right=right;
if(left==right)
{
return tree[root].Max=0;
}
int a,b,middle=(left+right)>>1;
a=create(root<<1,left,middle);
b=create(root<<1|1,middle+1,right);
return tree[root].Max=max(a,b);
}

void update_mark(int root)
{
if(tree[root].mark)
{
tree[root].Max+=tree[root].mark;
if(tree[root].left!=tree[root].right)
{
tree[root<<1].mark+=tree[root].mark;
tree[root<<1|1].mark+=tree[root].mark;
}
tree[root].mark=0;
}
}

int calculate(int root,int left ,int right)
{
update_mark(root);
if(tree[root].left>right||tree[root].right=left&&tree[root].right<=right)
{
return tree[root].Max;
}
int a,b;
a=calculate(root<<1,left,right);
b=calculate(root<<1|1,left,right);
return max(a,b);
}

int update(int root,int left,int right,int val)
{
update_mark(root);
if(tree[root].left>right||tree[root].right=left&&tree[root].right<=right)
{
tree[root].mark+=val;
update_mark(root);
return tree[root].Max;
}
int a=update(root<<1,left,right,val);
int b=update(root<<1|1,left,right,val);
return tree[root].Max=max(a,b);

}

int main()
{
int L;
while(scanf("%d",&L)!=EOF)
{
create(1,0,L);
int x,y,z;
while(scanf("%d%d%d",&x,&y,&z)!=EOF)
{
if(x>y)
swap(x,y);
if(x!=-1)
{
update(1,x,y,z);
}
else break;
}
int k=calculate(1,0,L);
int locl,locr;
for(int i=0; i<=L;i++)
{
if(calculate(1,i,i)==k)
{

            locl=i;  
            break;  
        }  
    }  
    for(int i=L;i>=0;i--)  
    {  
        if(calculate(1,i,i)==k)  
        {  
            locr=i;  
            break;  
        }  
    }  
    printf("%d,%d\\n",locl,locr);  
}  
return 0;  

}
/*
10
1 2 3
0 0 3
5 8 4
0 0 3
2 2 2
-1 1 3
*/

正确 代码：

#include  

struct Tree  

{  

    int left,right,cover;  

} tree[15000<<2];  
int covered=0;  
void create(int root,int left,int right)  
{  
    tree[root].left=left;  
    tree[root].right=right;  
    tree[root].cover=0;  
    if(right==left)  
        return ;  
    int mid=(left+right)>>1;  

    create(root<<1,left,mid);  

    create(root<<1|1,mid+1,right);  

}

void update(int root,int left,int right,int val)  

{  

    if(left<=tree[root].left&&tree[root].right<=right)  
    {  
        tree[root].cover+=val;  
        return ;  
    }  
    int m=(tree[root].left+tree[root].right)>>1;  

    if(m>=left)update(root<<1,left,right,val);  

    if(m<right)update(root<<1|1,left,right,val);  

}
void  calculate(int root,int x)  

{  

    covered+=tree[root].cover;  

    if(tree[root].left==tree[root].right)  

        return ;  

    int m=(tree[root].left+tree[root].right)>>1;  

    if(m>=x)  

        calculate(root<<1,x);  

    else  

        calculate(root<<1|1,x);
}
int main()  

{  

    int L;  

    while(scanf("%d",&L)!=EOF)  

    {  

        create(1,0,L);  

        int x,y,z;  

        while(scanf("%d%d%d",&x,&y,&z))  

        {  

            if(x==-1)  

                break;  

            update(1,x,y,z);  

        }  

        int loc1, loc2,Max=0;  

        for(int i=0; i<=L; i++)  
        {  
            covered=0;  
            calculate(1,i);  
            if(covered>Max)  

            {  

                Max=covered;  

                loc1=i;  

            }  

        }  

        for(int i=L,Max=0; i>=0; i--)  

        {  

            covered=0;  

            calculate(1,i);  

            if(covered>Max)  

            {  

                Max=covered;  

                loc2=i;  

            }  

        }  

        printf("%d %d\n",loc1,loc2);  

    }  

    return 0;  

}