HDU-2227 Find the nondecreasing subsequences 树状数组+DP

该题题义是求一个序列中非递减的子序列的个数，其实就是一个求逆序对的题，这里当然就是用树状数组来解决了。

首先对输入数据进行离散化，以便于在树状数组上面工作，然后利用DP公式计算ans[i] = sum{ ans[j], j < i }，可以理解为在前面的所有满足要求的集合上加上这个较大的数。
参看http://www.cppblog.com/menjitianya/archive/2011/04/06/143510.aspx

代码如下：

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <map>
#define MAXN 100005
#define MOD 1000000007
typedef long long ll;
using namespace std;

ll c[MAXN];
int N, val[MAXN], t[MAXN];

inline int lowbit( int x )
{
	return x & (-x);
}

inline void modify( int pos, ll val )
{
	for (int i = pos; i <= N; i += lowbit(i) )
	{
		c[i] += val;
		if (c[i] >= MOD)	c[i] %= MOD;
	}
}

inline ll sum( int pos )
{
	int s = 0;
	for (int i = pos; i > 0; i -= lowbit(i))
	{
		s += c[i];
		if (s >= MOD)	s %= MOD;
	}
	return s;
}

int main(  )
{
	while (scanf("%d", &N) == 1)
	{
		ll ans = 0;
		map<int,int>mp;
		memset(c, 0, sizeof(c));
		for (int i = 1; i <= N; ++i)
		{
			scanf("%d", &val[i]);
			t[i] = val[i];
		}
		sort(t + 1, t + N + 1);
		int cnt = unique(t + 1, t + N + 1) - t - 1;
		for (int i = 1; i <= cnt; ++i)
		{
			mp[t[i]] = i;
		}
		modify(1, 1);
		for (int i = 1; i <= N; ++i)
		{
			int x = sum(mp[val[i]]);
			ans += x;
			if (ans >= MOD)	ans %= MOD;
			modify(mp[val[i]], x);
		}
		printf("%lld\n", ans);
	}
	return 0;
}

以下是我的错误代码，套用了公式 2^N-1 （其中N为逆序对数，求解时-1就用自身的单点集合来中和了，所以直接就用2^N来计算了） 来计算。可是对于下面的情况无法给出正确的结果。

3

2 1 2

其结果会与

3

2 1 3

相同，多算了( 2, 1, 2 ) 这类集合......

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <map>
#define MAXN 100005
#define MOD 1000000007
typedef long long ll;
using namespace std;

ll c[MAXN];
int rec[MAXN];
int N, val[MAXN], t[MAXN];

inline int lowbit( int x )
{
	return x & (-x);
}

inline void modify( int pos, ll val )
{
	for (int i = pos; i <= N; i += lowbit(i) )
	{
		c[i] += val; 
	}
}

inline ll sum( int pos )
{
	int s = 0;
	for (int i = pos; i > 0; i -= lowbit(i))
	{
		s += c[i]; 
	}
	return s;
}

int main(  )
{
    rec[0] = 1;
    for (int i = 1; i < MAXN; ++i)
    {
        rec[i] = (rec[i-1]*2)%MOD;
    }
	while (scanf("%d", &N) == 1)
	{
		ll ans = 0;
		map<int,int>mp;
		memset(c, 0, sizeof(c));
		for (int i = 1; i <= N; ++i)
		{
			scanf("%d", &val[i]);
			t[i] = val[i];
		}
		sort(t + 1, t + N + 1);
		int cnt = unique(t + 1, t + N + 1) - t - 1;
		for (int i = 1; i <= cnt; ++i)
		{
			mp[t[i]] = i;
		}
		for (int i = 1; i <= N; ++i)
		{
			int x = sum(mp[val[i]]); 
		    ans += rec[x];
			if (ans >= MOD)	ans %= MOD;
			modify(mp[val[i]], 1);
		}
		printf("%lld\n", ans);
	}
	return 0;
}