bzoj 2274 [Usaco2011 Feb]Generic Cow Protests

Description
Farmer John’s N (1 <= N <= 100,000) cows are lined up in a row and
numbered 1..N. The cows are conducting another one of their strange
protests, so each cow i is holding up a sign with an integer A_i
(-10,000 <= A_i <= 10,000).
FJ knows the mob of cows will behave if they are properly grouped
and thus would like to arrange the cows into one or more contiguous
groups so that every cow is in exactly one group and that every
group has a nonnegative sum.
Help him count the number of ways he can do this, modulo 1,000,000,009.
By way of example, if N = 4 and the cows’ signs are 2, 3, -3, and
1, then the following are the only four valid ways of arranging the
cows:
(2 3 -3 1)
(2 3 -3) (1)
(2) (3 -3 1)
(2) (3 -3) (1)
Note that this example demonstrates the rule for counting different
orders of the arrangements.
给出n个数，问有几种划分方案（不能改变数的位置），使得每组中数的和大于等于0。输出方案数除以 1000000009的余数。

Input
* Line 1: A single integer: N
* Lines 2..N + 1: Line i + 1 contains a single integer: A_i
Output
* Line 1: A single integer, the number of arrangements modulo 1,000,000,009.

Sample Input
4
2
3
-3
1
Smple Output
4

Solution

可得简单dp如下

    cin>>n;
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&a[i]);
        s[i]=s[i-1]+a[i];
    }
    f[0]=1;
    for(int i=1;i<=n;i++) 
    {
        for(int j=0;jif(s[i]-s[j]>=0) f[i]=(f[i]+f[j])%mod;
    }
    cout<

完了这是n^2的怎么办
其实很多题都是像这样的套路，关键就是对dp条件式的转化
s[i]-s[j]>=0即s[i]>=s[j]
我们可以离散化s数组，然后用线段树来求和。

#include
#include
#include
#include
#define ll long long
#define p1 id<<1
#define p2 id<<1^1
using namespace std;
const int mod=1000000009;
int n,x;
int a[100005],f[100005],tree[400005];
ll s[100005];
struct ty
{
    ll v;
    int id;
}q[100005];
bool cmp(ty x,ty y)
{
    return x.v<y.v;
}
int query(int id,int l,int r,int x,int y)
{
    if(x<=l&&r<=y) return tree[id];
    int mid=(l+r)/2;
    if(y<=mid) return query(p1,l,mid,x,y);
    else
    if(x>mid) return query(p2,mid+1,r,x,y);
    else return (query(p1,l,mid,x,mid)+query(p2,mid+1,r,mid+1,y))%mod;
}
void update(int id,int l,int r,int x,int y)
{
    if(l==r) 
    {
        tree[id]=(tree[id]+y)%mod;
        return;
    }
    int mid=(l+r)/2;
    if(x<=mid) update(p1,l,mid,x,y); else update(p2,mid+1,r,x,y);
    tree[id]=(tree[p1]+tree[p2])%mod;
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&a[i]);
        s[i]=s[i-1]+a[i];
    }
    for(int i=0;i<=n;i++) 
    {
        q[i].v=s[i];
        q[i].id=i;
    }
    sort(q,q+n+1,cmp);
    x=1;
    s[q[0].id]=x;
    for(int i=1;i<=n;i++) 
    {
        if(q[i].v!=q[i-1].v) x++;
        s[q[i].id]=x;
    }
    f[0]=1;
    update(1,1,x,s[0],f[0]);
    for(int i=1;i<=n;i++) 
    {
        f[i]=query(1,1,x,1,s[i]);
        update(1,1,x,s[i],f[i]);
    }
    cout<return 0;
}