2017 ACM-ICPC 亚洲区（南宁赛区）网络赛The Heaviest Non-decreasing Subsequence Problem（线段树优化DP）

Let SS be
a sequence of integers s_{1}s​1​​, s_{2}s​2​​, ......, s_{n}s​n​​ Each
integer is is associated with a weight by the following rules:
(1) If is is negative, then its weight is 00.

(2) If is is greater than or equal to 1000010000,
then its weight is 55.
Furthermore, the real integer value of s_{i}s​i​​ is s_{i}-10000s​i​​−10000 .
For example, if s_{i}s​i​​ is 1010110101,
then is is reset to 101101 and
its weight is 55.

(3) Otherwise, its weight is 11.

A non-decreasing subsequence of SS is
a subsequence s_{i1}s​i1​​, s_{i2}s​i2​​, ......, s_{ik}s​ik​​,
with i_{1}<i_{2}\
...\ <i_{k}i​1​​<i​2​​ ... <i​k​​,
such that, for all 1
\leq j<k1≤j<k,
we have s_{ij}<s_{ij+1}s​ij​​<s​ij+1​​.

A heaviest non-decreasing subsequence of SS is
a non-decreasing subsequence with the maximum sum of weights.

Write a program that reads a sequence of integers, and outputs the weight of its

heaviest non-decreasing subsequence. For example, given the following sequence:

8080 7575 7373 9393 7373 7373 1010110101 9797 -1−1 -1−1 114114 -1−1 1011310113 118118

The heaviest non-decreasing subsequence of the sequence is <73,
73, 73, 101, 113, 118><73,73,73,101,113,118> with
the total weight being 1+1+1+5+5+1
= 141+1+1+5+5+1=14.
Therefore, your program should output 1414 in
this example.

We guarantee that the length of the sequence does not exceed 2*10^{5}2∗10​5​​

Input Format

A list of integers separated by blanks:s_{1}s​1​​, s_{2}s​2​​,......,s_{n}s​n​​

Output Format

A positive integer that is the weight of the heaviest non-decreasing subsequence.

样例输入

80 75 73 93 73 73 10101 97 -1 -1 114 -1 10113 118

样例输出

14

题意：求递增序列最大权值和。（数的最大值不超过20001）

#include <cstdio>
#include<cstring>
#include <algorithm>
using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 22222;
int MAX[maxn << 2];
void PushUP(int rt) {
MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]);
}
void update(int p, int w,int l, int r, int rt) {
if (l == r) {
MAX[rt]=max(w,MAX[rt]);//dp方程:dp[i]=max(dp[i],dp[j]+w)
return;
}
int m = (l + r) >> 1;
if (p <= m) update(p,w, lson);
else update(p,w, rson);
PushUP(rt);
}
int query(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) {
return MAX[rt];
}
int m = (l + r) >> 1;
int ret = 0;
if (L <= m) ret = max(ret, query(L, R, lson));
if (R > m) ret = max(ret, query(L, R, rson));
return ret;
}
int main() {
memset(MAX, 0, sizeof(MAX));
int n, w;
while (scanf("%d", &n) != EOF)
{
if (n < 0) continue;
else if (n >= 10000)
{
w = 5;
n -= 10000;
}
else w = 1;
update(n, query(1,n,1,10001,1) + w,1,10001,1);//在n出现之前，值比n小的，最大权值
}
printf("%d\n", query(1,10001,1,10001,1));
return 0;
}