HDU - 2227 Find the nondecreasing subsequences (树状数组 + 子序列 + 离散化)

HDU - 2227

Find the nondecreasing subsequences

Time Limit: 5000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Submit Status

Description

How many nondecreasing subsequences can you find in the sequence S = {s1, s2, s3, ...., sn} ? For example, we assume that S = {1, 2, 3}, and you can find seven nondecreasing subsequences, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

 

Input

The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next line contains n integers {s1, s2, s3, ...., sn}, 1 <= n <= 100000, 0 <= si <= 2^31.

 

Output

For each test case, output one line containing the number of nondecreasing subsequences you can find from the sequence S, the answer should % 1000000007.

 

Sample Input

31 2 3 

 

Sample Output

7 

题意:给予一个序列求解所构成的不下降子序列的个数 

通过离散化，将0->2^32分散到0->1e5之中
到达j处的不下降子序列的个数dp[j] = sum{dp[i]} + 1 {i < j && S[i] <= S[j] && i > 0};

状态方程推导:

假设dp[j]为S[j]为结尾的序列的个数
那么if(S[j] < S[i])那么 dp[i] += dp[j]
为什么呢，举个简单的例子
如果以S[j]为结尾的序列是一下:
最原始的序列:S = { 1, 5, 3, 6, 7, 5, 2, 1, 8, 9};
如果j = 4的话
那么以S[j]为结尾的不下降子序列如下:
6
1 6
5 6
3 6
1 3 6
1 5 6
那么如果我们取i = 5时
S[i] >= S[j]可以得到
6 7
1 6 7
5 6 7
3 6 7
1 3 6 7
1 5 6 7
就是dp[i] += dp[j]如此将所有S[j] <= S[i]都加起来
然后加上本身7加上那么状态转移方程极为dp[i] = sum{dp[j]} + 1

如此计算sum{dp[j],i < j && S[i] <= S[j] && i > 0}便是我们的难题了

而求和问题解决的话，可以基本能分为三类：

一：树状数组

二：线段树

三：前缀和

此处则是用了树状数组求解满足条件的数据和

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;typedef long long LL;const int MAXN = 1e5 + 5;const LL mod = 1e9 + 7;int N, A[MAXN], S[MAXN];LL C[MAXN];int Crete(LL x) {    int lb = 0,ub = N;    while(ub - lb > 1) {        int mid = (lb + ub) >> 1;        if(A[mid] >= x) ub = mid;        else lb = mid;    }    return ub;}int lowbit(int x) {    return x & (-x);}void add(int x,LL c) {    while(x <= N) {        C[x] += c;        C[x] %= mod;        x += lowbit(x);    }}int sum(int x) {    LL ret = 0;    while(x > 0) {        ret += C[x];        ret %= mod;        x -= lowbit(x);    }    return ret;}int main() {    while(~ scanf("%d", &N)) {        for(int i = 1; i <= N; i ++) {            scanf("%d", &S[i]);            A[i] = S[i];        }        sort(A + 1, A + N + 1);        LL ans = 0;        memset(C, 0, sizeof(C));        for(int i = 1; i <= N; i ++) {            int ret = lower_bound(A + 1, A + N + 1, S[i]) - A;            //int ret = Crete(S[i]);            int res = sum(ret) + 1;            ans += res;            ans %= mod;            add(ret, res);        }        printf("%I64d\n", ans % mod);    }    return 0;}