ICPC2017网络赛（南宁）子序列最大权值（树状数组+dp）

https://nanti.jisuanke.com/t/17319

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
题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

【题意】：

给出一组数列，负数的价值为0,，>=10000的价值为5，但值要减掉1万，其余的数价值为1；

问，选一个不减子序列，使得价值和最大，并输出这个价值

【分析】：

每加入一个新元素，用前面的，小数，最大权值加上当前元素的权值，作为当前元素的最优权值。

别的队的同学用拆点，把权值为5的点拆成5个1，跑一边最长不减子序列就可以

【代码】：

[cpp] view plain copy

1. #include <stdio.h>  
2. #include <math.h>  
3. #include <stdlib.h>  
4. #include <string.h>  
5. #include <time.h>  
6. #include <iostream>  
7. #include <algorithm>  
8. #include <string>  
9. #include <queue>  
10. #include <stack>  
11. #include <vector>  
12. #include <set>  
13. #include <list>  
14. #include <map>  
15. #define mset(a,i) memset(a,i,sizeof(a))  
16. using namespace std;  
17. typedef long long ll;  
18. const int INF=0x3f3f3f3f;  
19. const int MAX=50005;  
20. struct tree{  
21.     int n;  
22.     int a[MAX],Max[MAX],Min[MAX];//Max维护最大值 ,Min最小  
23.     void init(int N)  
24.     {  
25.         n=N;  
26.         for(int i=0;i<=N;i++)  
27.             Max[i]=-(Min[i]=INF);  
28.     }  
29.     void update(int x,int num)//单点更新  
30.     {  
31.         a[x]=num;  
32.         while(x<=n)  
33.         {  
34.             Min[x]=Max[x]=a[x];  
35.             int lx=x&-x;  
36.             for(int i=1;i<lx;i<<=1)  
37.             {  
38.                 Max[x]=max(Max[x],Max[x-i]);  
39.                 Min[x]=min(Min[x],Min[x-i]);  
40.             }  
41.             x+=x&-x;  
42.         }  
43.     }  
44.     int Qmax(int x,int y)//[x,y]最大值  
45.     {  
46.         int ans=-INF;  
47.         while(y>=x)  
48.         {  
49.             ans=max(a[y], ans);  
50.             y--;  
51.             for(;y-(y&-y)>=x;y-=(y&-y))  
52.                 ans=max(Max[y],ans);  
53.         }  
54.         return ans;  
55.     }  
56.     int Qmin(int x, int y)  
57.     {  
58.         int ans=INF;  
59.         while(y>=x)  
60.         {  
61.             ans=min(a[y],ans);  
62.             y--;  
63.             for(;y-(y&-y)>= x; y-=(y&-y)  )  
64.                 ans=min(Min[y],ans);  
65.         }  
66.         return ans;  
67.     }  
68. }C;  
69. int a[303030];  
70. int w[303030];  
71. int main()  
72. {  
73.     int n=1;  
74.     while(~scanf("%d",&a[n]))n++;n--;  
75.     int top=0;  
76.     for(int i=1;i<=n;i++)  
77.     {  
78.         if(a[i]<0)w[i]=0;  
79.         else if(a[i]>10000){  
80.             a[i]-=10000;  
81.             w[i]=5;  
82.         }  
83.         else w[i]=1;  
84.         a[i]+=21000;  
85.         top=max(top,a[i]);  
86.     }  
87.     C.init(top);  
88.     for(int i=1;i<=n;i++)  
89.     {  
90.         int dp=w[i]+C.Qmax(1,a[i]);//查询小的数的最大权值和  
91.         C.update(a[i],dp);//更新a[i]时的最大值  
92.     }  
93.     printf("%d\n",C.Qmax(1,top));  
94.     return 0;  
95. }