Hdu 5147 Sequence II(树状数字 or 线段树 + 输入外挂 前缀和+后缀和)

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题意:

给定1~n的一个排列 用A[]数组保存,问有多少下标(a,b,c,d)四元组满足:
a < b < c < d 且 A[a] < A[b],A[c] < A[d]

解析:

题目中n的范围是50000,O(n^2) 复杂度肯定超时。那么这题明显考察的是log2(n)的算法,对于这题可以用线段树或者树状数组,同时要用到输入外挂,不然会超时。

思路(参考别人做法)

枚举c的位置,那么每一次枚举中的方法数为 1~c-1 中(a,b)的个数 乘以 c~n中(c,d)的个数。累加起来即为答案。

1~c-1中(a,b)的个数相当于枚举b的位置,然后计算出b前面有多少数比A[b]小,该值要保存下来,下一次枚举c的时候,该值再加上c-1前面有多少比a[c-1]小的数即为当前情况下1~c-1中(a,b)的个数,也就是b=c-1的时候,因为枚举b之前的情况已经算过了。

因为当算c时只是算出 比c小的方法数,这个只是满足比c小的二元组的个数。而c前面的二元组也要计算在内,所以要加上c前面的比c小的个数。

用树状数组来做。本题n范围50000,而且每个数都不相同很关键。所以我们就开辟n个位置,一开始每个位置都是0,其实每个位置不是0就是1,因为每个数只有一个。

比如 1 3 2 4 5
一开始 c数组 0 0 0 0 0
先统计,再输入,因为计算a[i]前面有多少比它小的数,不包括它自己,而树状数组计算和的时候,要包括它自己。
i=1,树状数组求和前缀和 pre[1]=0 ,
此时0 0 0 0 0, 输入1,变为 1 0 0 0 0
i=2,a[2]=3,要看 3前面有多少个数,也就是看c数组的3个位置前面有多少个1,1代表已经输入,
发现1 0 0 0 0前三个数只有一个1,也就是pre[2]=1 (输入的第二个数之前只有1个比它小的),输入3以后,c数组变为 1 0 1 0 0
i=3, a[3]= 2, 要看2前面有多少个数,也就是看c数组前2个位置前面有多少个1,发现10100前两个数中只有一个1,也就是pre[3]=1

前缀和可以算出来,那么后缀和 = num2[i] = n-i-a[i]+num[i]+1;

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 5e4 + 10;
int n;
ll C[N], a[N];
inline void read(ll &x) {
    int flag = 0;
    x = 0;
    char c = getchar();
    if(c == '-')
        flag = 1;
    while(c < '0' || c > '9') {
        if(c == '-')
            flag = 1;
        c = getchar();
    }
    while(c >= '0' && c <= '9')
        x = x * 10 + c - '0', c = getchar();
    if(flag) x = -x;
}
inline int lowbit(int x) {
    return x & (-x);
}
int query(int x) {
    int ret = 0;
    while(x > 0) {
        ret += C[x];
        x -= lowbit(x);
    }
    return ret;
}
void add(int x, int d) {
    while(x <= n) {
        C[x] += d;
        x += lowbit(x);
    }
}
int num[N] , num2[N];
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        ll ans = 0, pre = 0;
        scanf("%d", &n);
        memset(C, 0, sizeof(C));
        for(int i = 1; i <= n; i++) {
            read(a[i]);
            num[i] = query(a[i]); //计算a[i]之前比a[i]小的数字
            num2[i] = n-i-a[i]+num[i]+1; //计算a[i]之后比a[i]大的数字
            add(a[i], 1);
            ans += pre * num2[i];
            pre += num[i];
        }
        printf("%lld\n", ans);
    }
    return 0;
}
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Sequence II 题意: 在数组中找到符合这个的条件的数的的个数: 1. 1≤abcd≤n 2. AaAb   3. AcAd 思路:对一个在位置i的数,用树状数组就能知道前面比它小的数,类似地,找到后面比他小地数; 根据后面比他小的数这个进行打表记录在这个位置后能够符合的对数; 相乘一下就行了; #include using namespa

Everything can be done! 311

[HDU 5147] Sequence II (树状数组+前缀和)

链接HDU 5147题意给出一个1到n的全排列,问有多少个四元组(a, b, c, d)满足a < b < c < d,且q[a] < q[b]、q[c] < q[d]。题解对每个位置记录其前方比它大的有多少个left[],其后方比它大的有多少个right[],这个利用树状数组逐个插入元素很容易做到。 直接枚举的话复杂度就还是O(n^2),需要记录left的前缀和sum,然后枚举right[i]

mysterious 571

HDU 5147 Sequence II 树状数组

Sequence II Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 375    Accepted Submission(s): 173 Problem Description Long long ago, there is

小人物的草稿本 672

hdu 5147 Sequence II

hdu 5147 Sequence II 树状数组 参考 http://blog.youkuaiyun.com/ljd4305/article/details/10101535 http://www.cnblogs.com/zhj5chengfeng/archive/2013/03/20/2965833.html 遇到了这道题不会,看了题解现学的树状数组= =望轻喷。题目大意:给定1~n的一个排列

MOMOXIRO的博客 499

HDU 5147 Sequence II (线段树求前缀和、后缀和)

Problem Description Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence. Pleas

SHUI 478

hdu 5147 树状数组+前缀和

题目: 题目分析:要保证两对数满足前面的大于后面的,第一对数的下标中较大的小于第二对数中下标较大的,那么我们可以利用前缀和和后缀和的思想,先预处理算出每一个数的前缀中比它小的数的个数,再计算出后缀中比它大的个数,利用树状数组可以快速求取,然后对前面的数取前缀和,也就是以它为较大数的所有数对的个数,枚举比它大的数中后缀的个数,相乘就是结果,具体实现详见代码: 代码如下

llin-黎辰 573
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