/****************************************************************************************************
尼瑪。。。一句話沒加能WA一個禮拜。。。線段樹+區間合併,主要是區間合併的細節問題。。。
****************************************************************************************************/
#include <iostream>
#include <functional>
#include <algorithm>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <sstream>
#include <utility>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <limits>
#include <memory>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;
#define LOWBIT(x) ( (x) & ( (x) ^ ( (x) - 1 ) ) )
#define CLR(x, k) memset((x), (k), sizeof(x))
#define CPY(t, s) memcpy((t), (s), sizeof(s))
#define SC(t, s) static_cast<t>(s)
#define LEN(s) static_cast<int>( strlen((s)) )
#define SZ(s) static_cast<int>( (s).size() )
typedef double LF;
//typedef long long LL; //GNU C++
//typedef unsigned long long ULL;
typedef __int64 LL; //Visual C++ 2010
typedef unsigned __int64 ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;
typedef map<int, int>::iterator MI;
typedef vector<int>::iterator VI;
typedef list<int>::iterator LI;
typedef set<int>::iterator SI;
template <typename T>
T sqa(const T &x)
{
return x * x;
}
template <typename T>
T gcd(T a, T b)
{
if (!a || !b)
{
return max(a, b);
}
T t;
while (t = a % b)
{
a = b;
b = t;
}
return b;
}
template <typename T>
T ext_gcd(T a, T b, T &x, T &y)
{
if (!b)
{
x = 1;
y = 0;
return a;
}
T d = ext_gcd(b, a % b, x, y);
T t = x;
x = y;
y = t - a / b * y;
return d;
}
template <typename T>
T invmod(T a, T p)
{
LL inv, y;
ext_gcd(a, p, inv, y);
inv < 0 ? inv += p : 0;
return inv;
}
const int INF_INT = 0x3f3f3f3f;
const LL INF_LL = 0x7fffffffffffffffLL; //15f
const double oo = 10e9;
const double eps = 10e-7;
const double PI = acos(-1.0);
#define ONLINE_JUDGE
const int MAXN = 200004;
int test, n, m, arr[MAXN];
struct Node
{
int left, right;
int lmx, rmx, mmx;
int len()
{
return right - left + 1;
}
}st[MAXN << 2];
int inline ll(const int &x)
{
return x << 1;
}
int inline rr(const int &x)
{
return x << 1 | 1;
}
void initST()
{
CLR(st, 0);
return ;
}
void updateMX(int t)
{
st[t].lmx = st[ ll(t) ].lmx;
st[t].rmx = st[ rr(t )].rmx;
int &mx = st[t].mmx;
mx = 0; //就TMD這句話沒加。。。
if (arr[ st[ ll(t) ].right ] < arr[ st[ ll(t) ].right + 1 ]) //左右節點的銜接處,細節
{
if (st[ ll(t) ].len() == st[ ll(t) ].lmx)
{
st[t].lmx += st[ rr(t) ].lmx;
}
if (st[ rr(t) ].len() == st[ rr(t) ].rmx)
{
st[t].rmx += st[ ll(t) ].rmx;
}
mx = st[ ll(t) ].rmx + st[ rr(t) ].lmx;
}
mx = max( mx, max( st[t].lmx, st[t].rmx ) ); //顯然st[t].mmx必然是這幾個量中最大的值
mx = max( mx, max( st[ ll(t) ].mmx, st[ rr(t) ].mmx ) );
return ;
}
void buildST(int l, int r, int t)
{
st[t].left = l;
st[t].right = r;
if (l == r)
{
st[t].lmx = st[t].rmx = st[t].mmx = 1;
return ;
}
int mid = (l + r) >> 1;
buildST(l, mid, ll(t));
buildST(mid + 1, r, rr(t));
updateMX(t); //更新節點
return ;
}
void updateST(int l, int r, int t)
{
if (l <= st[t].left && st[t].right <= r)
{
return ;
}
int mid = st[ ll(t) ].right;
if (r <= mid)
{
updateST(l, r, ll(t));
}
else if (l > mid)
{
updateST(l, r, rr(t));
}
else
{
updateST(l, mid, ll(t));
updateST(mid + 1, r, rr(t));
}
updateMX(t); //更新節點
return ;
}
int queryST(int l, int r, int t)
{
if (l <= st[t].left && st[t].right <= r)
{
return st[t].mmx;
}
int mid = st[ ll(t) ].right;
if (r <= mid)
{
return queryST(l, r, ll(t));
}
else if (l > mid)
{
return queryST(l, r, rr(t));
}
int mx = max( queryST(l, mid, ll(t)), queryST(mid + 1, r, rr(t)) );
if (arr[mid] < arr[mid + 1]) //這個地方有trick。。。要當心左右子節點可以銜接的情況
{
mx = max( mx, min( st[ ll(t) ].rmx, mid - l + 1 ) + min( st[ rr(t) ].lmx, r - mid ) );
}
return mx;
}
void ace()
{
char op[2];
int a, b;
for (scanf("%d", &test); test--; )
{
scanf("%d %d", &n, &m);
for (int i = 0; i < n; ++i)
{
scanf("%d", arr + i);
}
initST();
buildST(0, n - 1, 1);
while (m--)
{
scanf("%s %d %d", op, &a, &b);
if ('U' == op[0])
{
arr[a] = b;
updateST(a, a, 1);
}
else
{
printf("%d\n", queryST(a, b, 1));
}
}
}
return ;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
ace();
return 0;
}
HDU 3308 線段樹+區間合併
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