JOJ 2442: Be efficient 线段树的做法

2442: Be efficient

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	876	147	Standard

Given a triangle and the length of its edges,we could calculate its area easily. But the task is complicated by that given lots of triangles and several intervals represented by two numbers i and j (which means from triangle i to triangle j both inclusive),how could we determine which triangle in interval holds the largest area as fast as we can?

Input

There are several cases .In each case,the first line is two integer K(0 < K < 50000) and N(0 < N < 1000). K represent the number of trangles. The following K lines each contains 3 positive integers (a,b,c) means length of three edges.If a,b and c cannot form a triangle,then the area is zero. The next following N lines each contains 2 positive integers means the begin and the end of intervals.

Output

three edges of the largest triangle in each given interval.If area of the largest triangle is 0,output "I don't know~!". Order of edges is the same as input.output a blank line after each cases.

Sample Input

6 4
3 4 5
4 5 6
7 4 5
1 2 3
11 22 20
7 7 7
1 2
3 4
4 4
1 6

Sample Output

4 5 6
7 4 5
I don't know~!
11 22 20

Note

If there are severl triangles with the same largest area, you should output the first one.

 

Problem Source: slp3

 

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This problem is used for contest: 95 

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时间很快，可内存占用比较多，还是可以的

#include<stdio.h>

#include<math.h>

#define LL(x)  (x<<1)

#define RR(x) (x<<1|1)

 

struct Seg_Tree

{

    int left,right,num;

    int calmid() { return (right+left)/2;}

}tree[120020];

 

struct node

{

    int x,y,z;

    double value;

} num[50010];

 

int build(int left,int right,int idx)

{

    tree[idx].left=left;

    tree[idx].right=right;

    if(left==right)

    {

         return tree[idx].num=left;

    }

    int mid=tree[idx].calmid();

    int x=build(left,mid,LL(idx));

    int y=build(mid+1,right,RR(idx));

    if(num[x].value>num[y].value) return tree[idx].num=x;

    else return tree[idx].num=y;

}

 

int query(int left,int right,int idx)

{

    if(left<=tree[idx].left && right>=tree[idx].right) return tree[idx].num;

    int mid=tree[idx].calmid();

    if(right<=mid)  return query(left,right,LL(idx));

    else if(left>mid)  return query(left,right,RR(idx));

    else

    {

        int x=query(left,mid,LL(idx));

        int y=query(mid+1,right,RR(idx));

        if(num[x].value>num[y].value)   return x;

        else return y;

    }

}

int fabs(int c)

{

    if(c<0) return -c;

    return c;

}    

 

bool check(int a,int b,int c)

{

    if( a+b>c && a+c>b && b+c>a && fabs(a-b)<c && fabs(a-c)<b && fabs(b-c)<a)  return true;

    return  false;

}

double triangle(int a,int b,int c)

{

    double p=(a+b+c)*1.0/2;

    double s=sqrt(p*(p-a)*(p-b)*(p-c));

    return s;

}

int main()

{

    int k,n,a,b;

    while(scanf("%d%d",&k,&n)!=EOF)

    {

        for(int i=1;i<=k;i++)

        {

           scanf("%d%d%d",&num[i].x,&num[i].y,&num[i].z);

           if(check(num[i].x,num[i].y,num[i].z))

                   num[i].value=triangle(num[i].x,num[i].y,num[i].z);

           else  num[i].value=0;

        }

        build(1,k,1);

        for(int i=1;i<=n;i++)

        {

            scanf("%d%d",&a,&b);

            int xx=query(a,b,1);

            if(num[xx].value!=0)   printf("%d %d %d/n",num[xx].x,num[xx].y,num[xx].z);

            else printf("I don't know~!/n");

        }

        printf("/n");

    }

    return 0;

}