LIS again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 132 Accepted Submission(s): 34
Problem Description
a1<a2<…<aN. Let the subsequence of the given numeric sequence (
a1,a2,…,aN) be any sequence (
ai1,ai2,…,aiK), where
1≤i1<i2<…<iK≤N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and many others.
S[ i , j ] indicates (
ai,ai+1,ai+2,…,aj) .
Your program, when given the numeric sequence (
a1,a2,…,aN), must find the number of pair ( i, j) which makes the length of the longest ordered subsequence of S[ i , j ] equals to the length of the longest ordered subsequence of (
a1,a2,…,aN).
Input
a1,a2,…,aN separated by exact one space.
Process to the end of file.
[Technical Specification]
1≤n≤100000
0≤ai≤1000000000
Output
For each case，.output the answer in a single line.
Sample Input
3 1 2 3 2 2 1
Sample Output
1 3
Source
BestCoder Round #21
Recommend
heyang | We have carefully selected several similar problems for you: 5140 5139 5138 5137 5136
线段树+dp，dp[i]表示以第i个元素结尾的LIS的长度，线段树维护此LIS的最靠右的起始位置和长度，靠右是是为了完全统计出这样的区间，最后只要O(n)的时间去统计
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
const int inf = -0x3f3f3f3f;
int xis[N];
int arr[N];
int dp[N];
int sta[N];
int end[N];
int ans, s;
int cnt;
struct node
{
int l, r;
int val, s;
}tree[N << 2];
void build(int p, int l, int r)
{
tree[p].l = l;
tree[p].r = r;
tree[p].val = inf;
tree[p].s = inf;
if (l == r)
{
return;
}
int mid = (l + r) >>1;
build(p << 1, l, mid);
build(p << 1 | 1, mid + 1, r);
}
int BinSearch(int val)
{
int l = 1, r = cnt, mid;
int ans;
while (l <= r)
{
mid = (l + r) >> 1;
if (xis[mid] > val)
{
r = mid - 1;
}
else if (xis[mid] < val)
{
l = mid + 1;
}
else
{
ans = mid;
break;
}
}
return ans;
}
void update(int p, int pos, int val, int s)
{
if (tree[p].l == tree[p].r)
{
if (tree[p].val == val && tree[p].s < s)
{
tree[p].s = s;
}
else if (tree[p].val < val)
{
tree[p].val = val;
tree[p].s = s;
}
return;
}
int mid = (tree[p].l + tree[p].r) >> 1;
if (pos <= mid)
{
update(p << 1, pos, val, s);
}
else
{
update(p << 1 | 1, pos, val, s);
}
if (tree[p << 1].val > tree[p << 1 | 1].val)
{
tree[p].val = tree[p << 1].val;
tree[p].s = tree[p << 1].s;
}
else if (tree[p << 1].val < tree[p << 1 | 1].val)
{
tree[p].val = tree[p << 1 | 1].val;
tree[p].s = tree[p << 1 | 1].s;
}
else
{
tree[p].s = max(tree[p << 1].s, tree[p << 1 | 1].s);
}
}
void query(int p, int l, int r)
{
if (tree[p].l >= l && tree[p].r <= r)
{
if (tree[p].val > ans)
{
ans = tree[p].val;
s = tree[p].s;
}
else if (tree[p].val == ans && s < tree[p].s)
{
s = tree[p].s;
}
return;
}
int mid = (tree[p].l + tree[p].r) >> 1;
if (r <= mid)
{
query(p << 1, l, r);
}
else if (l > mid)
{
query(p << 1 | 1, l, r);
}
else
{
query(p << 1, l, mid);
query(p << 1 | 1, mid + 1, r);
}
}
int main()
{
int n;
while (~scanf("%d", &n))
{
__int64 ret = 0;
cnt = 0;
int tmp = 0;
for (int i = 1; i <= n; ++i)
{
scanf("%d", &arr[i]);
xis[++cnt] = arr[i];
}
if (n == 1)
{
printf("1\n");
continue;
}
sort (xis + 1, xis + cnt + 1);
cnt = unique(xis + 1, xis + cnt + 1) - xis - 1;
dp[1] = 1;
sta[1] = 1;
build(1, 1, cnt);
update(1, BinSearch(arr[1]), dp[1], sta[1]);
for (int i = 2; i <= n; ++i)
{
ans = inf;
s = inf;
if (BinSearch(arr[i]) == 1)
{
dp[i] = 1;
sta[i] = i;
}
else
{
query(1, 1, BinSearch(arr[i]) - 1);
}
if (ans == inf)
{
dp[i] = 1;
sta[i] = i;
}
else
{
dp[i] = ans + 1;
sta[i] = s;
}
update(1, BinSearch(arr[i]), dp[i], sta[i]);
tmp = max(tmp, dp[i]);
}
int last = n + 1;
for (int i = n; i >= 1; --i)
{
if (dp[i] == tmp)
{
end[i] = last - 1;
last = i;
}
}
for (int i = 1; i <= n; ++i)
{
if (dp[i] == tmp)
{
ret += (__int64)sta[i] * (__int64)(end[i] - i + 1);
}
}
printf("%I64d\n", ret);
}
return 0;
}
