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HDU_1394_线段数

http://acm.hdu.edu.cn/showproblem.php?pid=1394

 

线段数入门题,每次读入一个数,就寻找在树中比它大的值的个数,然后跟新数,把个数相加就是逆序数,每移动一个数,相当于当前逆序数加上比首元素大的数的数量,减去比首元素小的数的数量。

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct segtree{    int left,right,sum;}tree[10050];int n,a[5050];void build(int pos,int l,int r){    tree[pos].left = l;    tree[pos].right = r;    tree[pos].sum = 0;    if(l < r)    {        int mid = (l+r)/2;        build(pos*2,l,mid);        build(pos*2+1,mid+1,r);    }}int getsum(int pos,int l,int r){    if(tree[pos].left == l && tree[pos].right == r)   return tree[pos].sum;    int mid = (tree[pos].left+tree[pos].right)/2;    if(r <= mid)    return getsum(pos*2,l,r);    if(l > mid)     return getsum(pos*2+1,l,r);    return getsum(pos*2,l,mid)+getsum(pos*2+1,mid+1,r);}void update(int pos,int num){    tree[pos].sum++;    if(tree[pos].left == tree[pos].right)   return;    int mid = (tree[pos].left+tree[pos].right)/2;    if(num <= mid)  update(pos*2,num);    else    update(pos*2+1,num);}int main(){    while(~scanf("%d",&n))    {        int ans = 0;        build(1,0,n-1);        for(int i = 0;i < n;i++)        {            scanf("%d",&a[i]);            ans += getsum(1,a[i],n-1);            update(1,a[i]);        }        int temp = ans;        for(int i = 0;i < n-1;i++)        {            temp += n-1-a[i]-a[i];            ans = min(temp,ans);        }        printf("%d\n",ans);    }    return 0;}

 

HDU_1394_线段数