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 3 1 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; }