UVA 1471 Defense Lines

After the last war devastated your country, you - as the king of the land of Ardenia - decided it was

high time to improve the defense of your capital city. A part of your fortication is a line of mage

towers, starting near the city and continuing to the northern woods. Your advisors determined that the

quality of the defense depended only on one factor: the length of a longest contiguous tower sequence

of increasing heights. (They gave you a lengthy explanation, but the only thing you understood was

that it had something to do with ring energy bolts at enemy forces).

After some hard negotiations, it appeared that building new towers is out of question. Mages of

Ardenia have agreed to demolish some of their towers, though. You may demolish arbitrary number of

towers, but the mages enforced one condition: these towers have to be consecutive.

For example, if the heights of towers were, respectively, 5, 3, 4, 9, 2, 8, 6, 7, 1, then by demolishing

towers of heights 9, 2, and 8, the longest increasing sequence of consecutive towers is 3, 4, 6, 7.

Input

The input contains several test cases. The rst line of the input contains a positive integer

Z



25,

denoting the number of test cases. Then

Z

test cases follow, each conforming to the format described

below.

The input instance consists of two lines. The rst one contains one positive integer

n



2




10

5

denoting the number of towers. The second line contains

n

positive integers not larger than 10

9

separated by single spaces being the heights of the towers.

Output

For each test case, your program has to write an output conforming to the format described below.

You should output one line containing the length of a longest increasing sequence of consecutive

towers, achievable by demolishing some consecutive towers or no tower at all.

SampleInput

2

9

5 3 4 9 2 8 6 7 1

7

1 2 3 10 4 5 6

Output

4

6

解题思路：首先对数据进行离散化，然后用线段树维护一下即可，水之～～

#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <numeric>
#include <algorithm>
#include <functional>
using namespace std;
const int maxn = 2 * 100010;
int arr1[maxn], arr2[maxn];
int L[maxn], R[maxn];
int T, n, nn;

int bsearch(int x) {
int l = 0, r = nn - 1;
int mid;
while(l <= r) {
mid = (l + r) / 2;
if(arr1[mid] == x) {
return mid;
} else if(arr1[mid] < x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return -1;
}

int maxnum[maxn<<2];

void push_up(int rt) {
maxnum[rt] = max(maxnum[rt<<1], maxnum[rt<<1|1]);
}

void build(int l, int r, int rt) {
maxnum[rt] = 0;
if(l == r) return ;
int m = (l + r) >> 1;
build(l, m, rt<<1);
build(m + 1, r, rt<<1|1);
push_up(rt); ///
return ;
}

void update(int p, int x, int l, int r, int rt) {
if(l == r) {
maxnum[rt] = max(maxnum[rt], x);
return ;
}
/// push_down
int m = (l + r) >> 1;
if(p <= m) {
update(p, x, l, m, rt<<1);
} else {
update(p, x, m + 1, r, rt<<1|1);
}
push_up(rt);  ///
return ;
}

int query(int L, int R, int l, int r, int rt) {
if(L == l && R == r) {
return maxnum[rt];
}
int m = (l + r) >> 1;
/// push_down
if(R <= m) {
return query(L, R, l, m, rt<<1);
} else if(L > m) {
return query(L, R, m + 1, r, rt<<1|1);
} else {
int t1 = query(L, m, l, m, rt<<1);
int t2 = query(m + 1, R, m + 1, r, rt<<1|1);
return max(t1, t2);
}
}

int main() {

//freopen("aa.in", "r", stdin);

scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(int i = 0; i < n; ++i) {
scanf("%d", &arr1[i]);
arr2[i] = arr1[i];
}
sort(arr1, arr1 + n);
nn = unique(arr1, arr1 + n) - arr1;
for(int i = 0; i < n; ++i) {
arr2[i] = bsearch(arr2[i]) + 1;
}
memset(L, 0, sizeof(L));
memset(R, 0, sizeof(R));
L[0] = 1;
R[n-1] = 1;
for(int i = 1; i < n; ++i) {
if(arr2[i] > arr2[i-1]) {
L[i] = L[i-1] + 1;
} else {
L[i] = 1;
}
}
for(int i = n - 2; i >= 0; --i) {
if(arr2[i] < arr2[i+1]) {
R[i] = R[i+1] + 1;
} else {
R[i] = 1;
}
}
/*
for(int i = 0; i < n; ++i) {
printf("%d ", L[i]);
}
printf("\n");
for(int i = 0; i < n; ++i) {
printf("%d ", R[i]);
}
printf("\n");
*/
int ans = R[0];
build(1, nn, 1);
update(arr2[0], L[0], 1, nn, 1);
for(int i = 1; i < n; ++i) {
int t = 0;
if(arr2[i] - 1 > 0) {
t = query(1, arr2[i] - 1, 1, nn, 1);
}
ans = max(ans, t + R[i]);
update(arr2[i], L[i], 1, nn, 1);
}
printf("%d\n", ans);
}
return 0;
}