【bzoj2274】[Usaco2011 Feb]Generic Cow Protests dp+树状数组

题目描述

Farmer John's N (1 <= N <= 100,000) cows are lined up in a row andnumbered 1..N. The cows are conducting another one of their strangeprotests, 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 groupedand thus would like to arrange the cows into one or more contiguousgroups 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, and1, 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。

输入

* Line 1: A single integer: N
* Lines 2..N + 1: Line i + 1 contains a single integer: A_i

输出

* Line 1: A single integer, the number of arrangements modulo 1,000,000,009.

样例输入

4
2
3
-3
1

样例输出

4

---

题解

dp+树状数组

设dp[i]为前i个数的划分方案数。

则易推出dp[i]=∑dp[j](sum[j]≤sum[i],j<i)。

那么可以用树状数组维护sum[j]在区间内的dp[j]的和。

由于sum过大且可能出现非正数，所以要先将sum离散化。

#include <cstdio>
#include <algorithm>
#define MOD 1000000009
using namespace std;
struct data
{
	int sum , p;
}a[100010];
int f[100010] , dp[100010] , v[100010] , top;
bool cmp1(data a , data b)
{
	return a.sum < b.sum;
}
bool cmp2(data a , data b)
{
	return a.p < b.p;
}
void add(int p , int x)
{
	int i;
	for(i = p ; i <= top ; i += i & (-i))
		f[i] = (f[i] + x) % MOD;
}
int query(int p)
{
	int i , ans = 0;
	for(i = p ; i ; i -= i & (-i))
		ans = (ans + f[i]) % MOD;
	return ans;
}
int main()
{
	int n , i , t;
	scanf("%d" , &n);
	for(i = 1 ; i <= n ; i ++ )
		scanf("%d" , &t) , a[i].sum = a[i - 1].sum + t , a[i].p = i;
	sort(a , a + n + 1 , cmp1);
	v[0] = 0x80000000;
	for(i = 0 ; i <= n ; i ++ )
	{
		if(a[i].sum != v[top]) v[++top] = a[i].sum;
		a[i].sum = top;
	}
	sort(a , a + n + 1 , cmp2);
	dp[0] = 1;
	add(a[0].sum , 1);
	for(i = 1 ; i <= n ; i ++ )
		dp[i] = query(a[i].sum) , add(a[i].sum , dp[i]);
	printf("%d\n" , dp[n]);
	return 0;
}