poj3368线段树

题意:

给你一串上升序的数;

然后求i到j之间的数的最大重复次数;

理解:

这题稍微想想就是线段树了;

然而我第一次看见这个的时候还不会区间合并;

然后前一个周末练了这样的题型,于是今天再拿它练练;

这题就是注意合并区间;

代码如下:

我的线段树确实写的好复杂....

#include <cstdio>

#include <algorithm>
#include <vector>

using namespace std;

#define P pair<int, int>
#define Pi pair<P, P>
#define Px pair<int, Pi>
#define x first
#define y second

vector<int> vn(100010);
vector<Px> tree(500000);

Px build(int l, int r, int rx)
{
if (l == r)
{
return tree[rx] = Px(1, Pi(P(vn[l - 1], 1), P(vn[r - 1], 1)));
}

Px p1 = build(l, (l + r) / 2, rx * 2);
Px p2 = build((l + r) / 2 + 1, r, rx * 2 + 1);

if (p1.y.y.x == p2.y.x.x)
{
tree[rx].x = max(p1.x, max(p2.x, p1.y.y.y + p2.y.x.y));
if (p1.y.x == p1.y.y)
{
tree[rx].y.x = P(p1.y.y.x, p1.y.y.y + p2.y.x.y);
}
else tree[rx].y.x = p1.y.x;
if (p2.y.x == p2.y.y)
{
tree[rx].y.y = P(p2.y.x.x, p1.y.y.y + p2.y.x.y);
}
else tree[rx].y.y = p2.y.y;
}
else
{
tree[rx].x = max(p1.x, p2.x);
tree[rx].y.x = p1.y.x;
tree[rx].y.y = p2.y.y;
}
return tree[rx];
}

Px query(int l, int r, int L, int R, int rx)
{
if (l > R || r < L) return Px(0, Pi(P(0, 0), P(0, 0)));
if (l <= L && r >= R) return tree[rx];
Px p1 = query(l, r, L, (L + R) / 2, rx * 2);
Px p2 = query(l, r, (L + R) / 2 + 1, R, rx * 2 + 1);
//printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%d~~~\n", rx);
//printf("%d~~~%d~~~%d~~~%d~~~%d\n", p1.x, p1.y.x.x, p1.y.x.y, p1.y.y.x, p1.y.y.y);
//printf("%d~~~%d~~~%d~~~%d~~~%d\n", p2.x, p2.y.x.x, p2.y.x.y, p2.y.y.x, p2.y.y.y);

if (p1.x && p2.x)
{
Px p;
if (p1.y.y.x == p2.y.x.x)
{
p.x = max(p1.x, max(p2.x, p1.y.y.y + p2.y.x.y));
if (p1.y.x == p1.y.y)
{
p.y.x = P(p1.y.y.x, p1.y.y.y + p2.y.x.y);
}
else p.y.x = p1.y.x;
if (p2.y.x == p2.y.y)
{
p.y.y = P(p2.y.x.x, p1.y.y.y + p2.y.x.y);
}
else p.y.y = p2.y.y;
}
else
{
p.x = max(p1.x, p2.x);
p.y.x = p1.y.x;
p.y.y = p2.y.y;
}
return p;
}
else
{
if (p1.x) return p1;
if (p2.x) return p2;
return Px(0, Pi(P(0, 0), P(0, 0)));
}
}

int main()
{
int n, q;
while (scanf("%d", &n) && n)
{
scanf("%d", &q);
for (int i = 0; i < n; ++i)
{
scanf("%d", &vn[i]);
}
build(1, n, 1);
for (int i = 0; i < q; ++i)
{
int num1, num2;
scanf("%d%d", &num1, &num2);
printf("%d\n", query(num1, num2, 1, n, 1).x);
}
}

return 0;
}