HDU 3308 LCIS (线段树~)
2017-10-23 23:04
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这个题花了很长很长很长时间才解决,感觉理解线段树需要对递归有很好的掌握才行,至于这个题的做法,以后有时间我会补上,一时半会很难说清楚。我需要更多的练习才能完全掌握这些思想,BTW,os真是越看越烦,看不下去就开始写题,写完还得看TAT
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int maxn = 100000 + 5;
int msum[maxn << 2];
int lsum[maxn << 2];
int rsum[maxn << 2];
int num[maxn];
void pushup(int l, int r, int rt) {
lsum[rt] = lsum[2 * rt];
rsum[rt] = rsum[2 * rt + 1];
msum[rt] = max(msum[rt * 2], msum[rt * 2 + 1]);
int m = (l + r) >> 1;
int len = (r - l + 1);
if (num[m] < num[m + 1]) {
if (lsum[rt] == len - len / 2) lsum[rt] += lsum[2 * rt + 1];
if (rsum[rt] == len / 2) rsum[rt] += rsum[2 * rt];
msum[rt] = max(msum[rt], lsum[2 * rt + 1] + rsum[2 * rt]);
}
}
void update(int p, int l, int r, int rt) {
//只起到一个更新作用,修改值已经在主函数中修改了
if (l == r) return;
int m = (l + r) >> 1;
if (p <= m)
update(p, lson);
else
update(p, rson);
pushup(l, r, rt);
}
int query(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) return msum[rt];
int m = (l + r) >> 1;
if (R <= m) return query(L, R, lson);
if (L > m) return query(L, R, rson);
int a, b;
a = query(L, R, lson);
b = query(L, R, rson);
int ans = max(a, b);
if (num[m] < num[m + 1]) {
int c;
c = min(rsum[2 * rt], m - L + 1) + min(lsum[2 * rt + 1], R - m);
ans = max(c, ans);
}
return ans;
}
void build(int l, int r, int rt) {
if (l == r) {
lsum[rt] = rsum[rt] = msum[rt] = 1;
return;
}
else {
int m = (l + r) >> 1;
build(lson);
build(rson);
pushup(l, r, rt);
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i
4000
<= n; i++) {
scanf("%d", &num[i]);
}
build(1, n, 1);
while (m--) {
char op[121];
scanf("%s", op);
int a, b;
scanf("%d%d", &a, &b);
if (op[0] == 'U') {
a++;
num[a] = b;
update(a, 1, n, 1);
} else {
a++;
b++;
printf("%d\n", query(a, b, 1, n, 1));
}
}
}
}
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int maxn = 100000 + 5;
int msum[maxn << 2];
int lsum[maxn << 2];
int rsum[maxn << 2];
int num[maxn];
void pushup(int l, int r, int rt) {
lsum[rt] = lsum[2 * rt];
rsum[rt] = rsum[2 * rt + 1];
msum[rt] = max(msum[rt * 2], msum[rt * 2 + 1]);
int m = (l + r) >> 1;
int len = (r - l + 1);
if (num[m] < num[m + 1]) {
if (lsum[rt] == len - len / 2) lsum[rt] += lsum[2 * rt + 1];
if (rsum[rt] == len / 2) rsum[rt] += rsum[2 * rt];
msum[rt] = max(msum[rt], lsum[2 * rt + 1] + rsum[2 * rt]);
}
}
void update(int p, int l, int r, int rt) {
//只起到一个更新作用,修改值已经在主函数中修改了
if (l == r) return;
int m = (l + r) >> 1;
if (p <= m)
update(p, lson);
else
update(p, rson);
pushup(l, r, rt);
}
int query(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) return msum[rt];
int m = (l + r) >> 1;
if (R <= m) return query(L, R, lson);
if (L > m) return query(L, R, rson);
int a, b;
a = query(L, R, lson);
b = query(L, R, rson);
int ans = max(a, b);
if (num[m] < num[m + 1]) {
int c;
c = min(rsum[2 * rt], m - L + 1) + min(lsum[2 * rt + 1], R - m);
ans = max(c, ans);
}
return ans;
}
void build(int l, int r, int rt) {
if (l == r) {
lsum[rt] = rsum[rt] = msum[rt] = 1;
return;
}
else {
int m = (l + r) >> 1;
build(lson);
build(rson);
pushup(l, r, rt);
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i
4000
<= n; i++) {
scanf("%d", &num[i]);
}
build(1, n, 1);
while (m--) {
char op[121];
scanf("%s", op);
int a, b;
scanf("%d%d", &a, &b);
if (op[0] == 'U') {
a++;
num[a] = b;
update(a, 1, n, 1);
} else {
a++;
b++;
printf("%d\n", query(a, b, 1, n, 1));
}
}
}
}
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