基础线段树

 

Balanced Lineup

Time Limit:5000MS  Memory Limit:65536K
Total Submit:35 Accepted:22
Case Time Limit:2000MS

Description

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

 

 

 

 

6 31734251 54 62 2

 

 

 

 

Sample Output

 

 

 

 

630

写代码起来超快，看来寒假练的还是有点效果的，但致命的失误也是存在的，比如说写错了答案还是对的，所以自己造的小数据无论怎么样都是

直接跑出来的，这样就得重新检查代码了，也就是写起来快也就没用了。

贴第一次提交写错代码

#include<iostream>using namespace std; int n,q;const int N=200000;inline int max(int a,int b){return a>b?a:b;}inline int min(int a,int b){return a<b?a:b;}inline int mid(int a,int b){return (a+b)>>1;}struct seg_tree{    int l,r;    int mi,ma;};seg_tree dia[4*N]; int f[N+5]; void maketree(int l,int r,int index){    dia[index].l=l;    dia[index].r=r;    if(l==r)    {        dia[index].ma=dia[index].mi=f[l];        return ;    }    int midd=mid(l,r);    maketree(l,midd,index<<1);    maketree(midd+1,r,(index<<1)+1);    dia[index].ma=max(dia[index<<1].ma,dia[(index<<1)+1].ma);    dia[index].mi=min(dia[index<<1].mi,dia[(index<<1)+1].mi);} int finds_max(int l,int r,int c){    if(dia[c].l==dia[c].r)        return dia[c].ma;    if(dia[c].l==l&&dia[c].r==r)        return dia[c].ma;    int midd=mid(dia[c].l,dia[c].r);    if(r<=midd)        return finds_max(l,r,c*2);    else if(l>midd)        return finds_max(l,r,c*2+1);    else        return max(finds_max(l,midd,c*2),finds_max(l,r,c*2+1));//就是这里悲剧了。在右半部分写错了。应该是finds_max(midd+1,r,c*2+1)
} int finds_min(int l,int r,int c){    if(dia[c].l==dia[c].r)        return dia[c].mi;    if(dia[c].l==l&&dia[c].r==r)        return dia[c].mi;    int midd=mid(dia[c].l,dia[c].r);    if(r<=midd)        return finds_min(l,r,c*2);    else if(l>midd)        return finds_min(l,r,c*2+1);    else        return min(finds_min(l,midd,c*2),finds_min(l,r,c*2+1));//就是这里悲剧了。} int main(){    int i,j;    while(scanf("%d%d",&n,&q)!=EOF)    {        for(i=1;i<=n;i++)            scanf("%d",&f[i]);        maketree(1,n,1);        while(q--)        {            scanf("%d%d",&i,&j);            printf("%d/n",finds_max(i,j,1)-finds_min(i,j,1));        }    }    return 0;}

后来发现最大的和最小的可以放在一起找

又改了下，，其实错误是在改的时候发现的。

#include<iostream>

using namespace std;

int n,q;

const int N=50000;

const int inf=999999999;

inline int max(int a,int b){return a>b?a:b;}

inline int min(int a,int b){return a<b?a:b;}

inline int mid(int a,int b){return (a+b)>>1;}

struct seg_tree

{

int l,r;

int mi,ma;

};

seg_tree dia[4*N];

int _max,_min;

int f[N+5];

void maketree(int l,int r,int index)

{

dia[index].l=l;

dia[index].r=r;

if(l==r)

{

dia[index].ma=dia[index].mi=f[l];

return ;

}

int midd=mid(l,r);

maketree(l,midd,index<<1);

maketree(midd+1,r,(index<<1)+1);

dia[index].ma=max(dia[index<<1].ma,dia[(index<<1)+1].ma);

dia[index].mi=min(dia[index<<1].mi,dia[(index<<1)+1].mi);

}

void finds(int l,int r,int c)

{

if(dia[c].l==dia[c].r)

{

_max=max(dia[c].ma,_max);

_min=min(dia[c].mi,_min);

return ;

}

if(dia[c].l==l&&dia[c].r==r)

{

_max=max(dia[c].ma,_max);

_min=min(dia[c].mi,_min);

return ;

}

int midd=mid(dia[c].l,dia[c].r);

if(r<=midd)

finds(l,r,c*2);

else if(l>midd)

finds(l,r,c*2+1);

else

{

finds(l,midd,c*2);

finds(midd+1,r,c*2+1);

}

}

int main()

{

// freopen("1.in","r",stdin);

// freopen("1.out","w",stdout);

int i,j;

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

{

for(i=1;i<=n;i++)

scanf("%d",&f[i]);

maketree(1,n,1);

while(q--)

{

scanf("%d%d",&i,&j);

_max=0;

_min=inf;

finds(i,j,1);

printf("%d/n",_max-_min);

}

}

return 0;

}

这题最方便还是用RMQ做，也是最适合用RMQ做的！有机会回头用它来做做看