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;
}