HDU 3308【线段树--c++版区间合并，dp】

LCIS

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 952 Accepted Submission(s): 447



[align=left]Problem Description[/align]
Given n integers.

You have two operations:

U A B: replace the Ath number by B. (index counting from 0)

Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].

[align=left]Input[/align]
T in the first line, indicating the case number.

Each case starts with two integers n , m(0<n,m<=105).

The next line has n integers(0<=val<=105).

The next m lines each has an operation:

U A B(0<=A,n , 0<=B=105)

OR

Q A B(0<=A<=B< n).

[align=left]Output[/align]
For each Q, output the answer.

[align=left]Sample Input[/align]

1
10 10
7 7 3 3 5 9 9 8 1 8
Q 6 6
U 3 4
Q 0 1
Q 0 5
Q 4 7
Q 3 5
Q 0 2
Q 4 6
U 6 10
Q 0 9

[align=left]Sample Output[/align]

1
1
4
2
3
1
2
5

[align=left]Author[/align]
shǎ崽

[align=left]Source[/align]
HDOJ Monthly Contest – 2010.02.06

[align=left]Recommend[/align]
wxl

//题目：LCIS hdu 3308

//求任意长度的最长连续子序列，且随时会改变序列里面的值

#include <stdio.h>
#include <memory.h>
#include <iostream>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define Maxn 100100
using namespace std;

int node[Maxn];
struct point {
int Msum;//范围内的最大的长度
int lsum, rsum;//从最左端起最长的长度，从最右端起最长的长度
int lnum, rnum;//最左端的数字，最右端的数字
bool mark;//是否这一整段范围内的数都满足，即从左起的最大长度是整段的长度
void init() {
Msum = lsum = rsum = mark = 1;
}
void PushUp(point, point);
}pot[4 * Maxn];

inline int Max(int a, int b) {
if(a > b) return a;
return b;
}

void scan_d(int &a) {
char in;
in = getchar();
while( in != '-' && (in > '9' || in < '0')) {
in = getchar();
}
int mark = 0;
if (in =='-') {
mark = 1;
a = 0;
} else  a = in - '0';
while (in = getchar(), in >= '0' && in <= '9') {
a = a * 10 + in - '0';
}
if (mark) a = -a;
}

void point::PushUp(point ls, point rs) {//合并ls，rs两个区间
lsum = ls.lsum;
rsum = rs.rsum;
lnum = ls.lnum;
rnum = rs.rnum;
Msum = Max(ls.Msum, rs.Msum);
if (ls.rnum < rs.lnum) {
if(ls.mark&&rs.mark) {//如果左右的区间都是全的那么父亲区间也要取全
Msum = lsum = rsum = lsum + rsum;
mark = 1;
return ;
}
mark = 0;//注意
if (ls.mark) lsum += rs.lsum;
else if (rs.mark)  rsum += ls.rsum;
Msum = Max(Msum, rs.lsum + ls.rsum);
}
mark = 0;//注意
}

void Build(int l, int r, int rt) {
if (l == r) {
pot[rt].init();
pot[rt].lnum = pot[rt].rnum = node[l];//注意是node[l]
return ;
}
int m = (r + l) >> 1;
Build(lson);
Build(rson);
pot[rt].PushUp(pot[rt<<1], pot[rt<<1|1]);
}

void UpDate(int a, int b, int l, int r, int rt) {
if (l == r) {
pot[rt].lnum = pot[rt].rnum = b;
return ;
}
int m = (r + l) >> 1;
if (a <= m) UpDate(a, b, lson);
else UpDate(a, b, rson);
pot[rt].PushUp(pot[rt<<1], pot[rt<<1|1]);
}

point query(int L, int R, int l, int r, int rt) {
point ret;
if (L > r || R < l) {
ret.Msum = -1;
return ret;
}
if (L <= l && R >= r) {
return pot[rt];
}
int m = (r + l) >> 1;
point a = query(L, R, lson);//与c不一样的地方
point b = query(L, R, rson);
if (a.Msum == -1)  return b;
if (b.Msum == -1)  return a;
ret.PushUp(a, b);
return ret;
}

int main()
{
int casenumber, n, m, a, b;
scan_d(casenumber);
while (casenumber--) {
scan_d(n);
scan_d(m);
for (int i = 1; i <= n; i++) scan_d(node[i]);
Build(1, n, 1);
char str;
for (int i = 0; i < m; i ++) {
str = getchar();
scan_d(a);
scan_d(b);
if (str == 'U') {
UpDate(a + 1, b, 1, n, 1);//注意题目的范围是从0-n-1
}
else {
point h = query(a + 1, b + 1, 1, n, 1);
printf("%d\n", h.Msum);
}
}
}
return 0;
}