poj2761

题意：给出每只狗的pretty value，然后多次询问，每次输出区间[i,j](狗站成一排，从第i只到第j只)的第k小的值是多少。区间之间有交叉，但是没有完全包含。

分析：先把所有区间排序，然后从左至右把每个区间用线段树维护离散化后的pretty value，即线段树的区间的意义是pretty value。每次删除在上一个区间中且不在当前区间中的节点，插入在当前区间中且不在上一个区间中的节点，使得线段树中的节点恰好为该区间内的所有节点。然后查询第k个就容易了。

View Code

#include < iostream >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < algorithm >
using namespace std;

#define maxn 100005
#define maxm 50005

struct Interval
{
int s, e, k, pos, ans;
}interval[maxm];

struct Node
{
Node * pleft, * pright;
int l, r;
int num;
}tree[maxn * 3 ];

int dog[maxn], n, m, pretty[maxn], opdog[maxn], ncount;

bool operator < ( const Interval & a, const Interval & b)
{
if (a.s != b.s)
return a.s < b.s;
return a.e < b.e;
}

bool cmp( const Interval & a, const Interval & b)
{
return a.pos < b.pos;
}

int binarysearch( int a)
{
int l = 0 ;
int r = n - 1 ;
int mid = (l + r) / 2 ;
while (l < r)
{
if (pretty[mid] < a)
l = mid + 1 ;
else
r = mid;
mid = (l + r) / 2 ;
}
return l;
}

void buildtree(Node * proot, int l, int r)
{
proot -> l = l;
proot -> r = r;
proot -> num = 0 ;
if (l == r)
return ;
int mid = (l + r) / 2 ;
ncount ++ ;
proot -> pleft = tree + ncount;
ncount ++ ;
proot -> pright = tree + ncount;
buildtree(proot -> pleft, l, mid);
buildtree(proot -> pright, mid + 1 , r);
}

void ins(Node * proot, int a, int d)
{
proot -> num += d;
if (proot -> l == proot -> r)
return ;
int mid = (proot -> l + proot -> r) / 2 ;
if (a <= mid)
ins(proot -> pleft, a, d);
else
ins(proot -> pright, a, d);
}

int query(Node * proot, int k)
{
if (proot -> l == proot -> r)
return pretty[proot -> l];
if (proot -> pleft -> num >= k)
return query(proot -> pleft, k);
else
return query(proot -> pright, k - proot -> pleft -> num);
}

void input()
{
scanf( " %d%d " , & n, & m);
for ( int i = 0 ; i < n; i ++ )
{
scanf( " %d " , & dog[i]);
pretty[i] = dog[i];
}
for ( int i = 0 ; i < m ; i ++ )
{
scanf( " %d%d%d " , & interval[i].s, & interval[i].e, & interval[i].k);
interval[i].s -- ;
interval[i].e -- ;
interval[i].pos = i;
}
}

void work()
{
for ( int i = interval[ 0 ].s; i <= interval[ 0 ].e; i ++ )
ins(tree, dog[i], 1 );
interval[ 0 ].ans = query(tree, interval[ 0 ].k);
for ( int i = 1 ; i < m; i ++ )
{
for ( int j = interval[i - 1 ].s; j <= min(interval[i - 1 ].e, interval[i].s - 1 ); j ++ )
ins(tree, dog[j], - 1 );
for ( int j = max(interval[i - 1 ].e + 1 , interval[i].s); j <= interval[i].e; j ++ )
ins(tree, dog[j], 1 );
interval[i].ans = query(tree, interval[i].k);
}
}

int main()
{
// freopen("D:\\t.txt", "r", stdin);
input();
sort(interval, interval + m);
sort(pretty, pretty + n);
for ( int i = 0 ; i < n; i ++ )
{
dog[i] = binarysearch(dog[i]);
opdog[dog[i]] = i;
}
ncount = 0 ;
buildtree(tree, 0 , n - 1 );
work();
sort(interval, interval + m, cmp);
for ( int i = 0 ; i < m; i ++ )
printf( " %d\n " , interval[i].ans);
return 0 ;
}