poj1952(lis)

o(n^2)的做法
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll  num[9000];
int len[9000];
int n;
int v[9000];
int main()
{
while (scanf("%d", &n) != EOF)
{
int maxlen = 1;
for (int i = 0; i < n; i++)
{
scanf("%d", &v[i]);
len[i] = num[i] = 1;
}
for (int i = 1; i < n; i++)
{
ll tempnum = 1;
int tempmax = 1;
for (int j = i - 1; j >= 0; j--)
{
if (v[j] <v[i]||num[j]==0)continue;
if (v[j] == v[i])
{
num[j] = 0;
continue;
}
if ((len[j] + 1) > tempmax)
{
tempmax= len[j] + 1;
tempnum= num[j];
continue;
}
if ((len[j]+1)==tempmax)
{
tempnum += num[j];
}
}
len[i] = tempmax;
num[i] = tempnum;
maxlen = max(maxlen, len[i]);
}
int ans = 0;
for (int i = n-1; i >=0; i--)
if (len[i] == maxlen)
{
ans += num[i];
}
printf("%d %d\n", maxlen, ans);
}

}



由于常数的原因下面这个o(nlogn)的算法并没有快多少，不过对于n=300000也是可以处理的
当然也可以用树状数组优化此做法，但是只限于此题，若是n=300000就不能用树状数组了，要用set
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
typedef long long ll;
struct r1
{
	int v;
	ll num;
};
struct r2
{
	int v;
	ll realnum;
};
bool operator <(r1 a, r1 b)
{
	return a.v > b.v;
}
set<r1>s[6000];
set<r1>::iterator it;
r2 rank2[10000];
int l, r, n;
int v[10000];
ll sum[10000];
int find(int v)
{
	int templ = l;
	int tempr = r;
	while (templ< tempr)
	{
		int mid = (templ + tempr) / 2;
		if (rank2[mid].v <= v)
			tempr = mid;
		else
			templ = mid + 1;
	}
	if (rank2[tempr].v > v)
		tempr++;
	return tempr;
}
int main()
{
	while (scanf("%d", &n) != EOF)
	{
		l = r = 0;
		int pos;
		for (int i = 1; i <=n; i++)
		{
			scanf("%d", &v[i]);
			rank2[i].v = rank2[i].realnum = 0; sum[i] = 0;
			s[i].clear();
		}
		sum[0] = 1; rank2[0].v = 1000000000;
		for (int i = 1; i <= n; i++)
		{
			pos=find(v[i]);
			ll tempnum;
			r1 tempuse;
			tempuse.v = v[i];
			it=s[pos - 1].lower_bound(tempuse);
			if (it == s[pos - 1].end()||(it==(s[pos-1].end()--)))
			{
				tempnum = sum[pos - 1];
			}
			else
			{
				tempnum = sum[pos - 1] - it->num;
			}
			if (v[i] == rank2[pos].v)
			{
					if (tempnum > rank2[pos].realnum)
					{
						tempuse.v = v[i];
						tempuse.num = sum[pos];
						s[pos].erase(tempuse);
					}
				sum[pos] += (tempnum - rank2[pos].realnum);
			}
			else
			{
				sum[pos] += tempnum;
			}
			tempuse.num = sum[pos];
			tempuse.v = v[i];
			s[pos].insert(tempuse);
			rank2[pos].realnum = tempnum;
			rank2[pos].v = v[i];
			if (pos > r)
				r = pos;
		}
		printf("%d %d\n", r, sum[r]);
	}
	return 0;
}