2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 L. The Heaviest Non-decreasing Subsequence Problem（最长不下降子序列变形）_综合

Let $S$ be a sequence of integers $s_{1}$ , $s_{2}$ , $. . .$ , $s_{n}$ Each integer is is associated with a weight by the following rules:

(1) If is is negative, then its weight is $0$ .

(2) If is is greater than or equal to $10000$ , then its weight is $5$ . Furthermore, the real integer value of $s_{i}$ is $s_{i}-10000$ . For example, if $s_{i}$ is $10101$ , then is is reset to $101$ and its weight is $5$ .

(3) Otherwise, its weight is $1$ .

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

A heaviest non-decreasing subsequence of $S$ 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:

$80$ $75$ $73$ $93$ $73$ $73$ $10101$ $97$ $? 1$ $? 1$ $114$ $? 1$ $10113$ $118$

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

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

Input Format

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

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

题解：

把负数直接跳掉，把大于10000的取模后复制5次放进数列里面，其他的直接放进数列里面求最长不下降子序列

代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
#define ll long long
#define INF 1008611111
#define M (t[k].l+t[k].r)/2
#define lson k*2
#define rson k*2+1
using namespace std;
const int N =1e6+5;
int  a[N], f[N], d[N];  // f[i]用于记录 a[0...i]的最大长度
int bsearch(const int *f, int size, const int &a)
{int  l=0, r=size-1;while( l <= r ){int  mid = (l+r)>>1;if( a >= d[mid-1] && a < d[mid] )return mid;             // >&&<= 换为: >= && <else if( a <d[mid] )r = mid-1;else l = mid+1;}
}
int LIS(const int *a, const int &n)
{int  i, j, size = 1;d[0] = a[0]; f[0] = 1;for( i=1; i < n; ++i ){if( a[i] < d[0] )             // <= 换为: <j = 0;else if( a[i] >= d[size-1] ) // > 换为: >=j = size++;elsej = bsearch(d, size, a[i]);d[j] = a[i]; f[i] = j+1;}return size;
}
int main()
{int i,j,ans=0,d;while(scanf("%d",&d)!=EOF){if(d<0)continue;if(d>=10000){for(i=0;i<5;i++){a[ans]=d%10000;ans++;}}else{a[ans]=d;ans++;}}printf("%d\n",LIS(a,ans));return 0;
}