[SRM] 09 撕书狂魔CZL

Posted 2020-10-01 Lev今天学习了吗

A. 撕书Ⅰ

序列型DP。DP[i]表示当前编号结点的撕书页数。

那么我们有 DP[ i ] = DP[ i - y - 1 ] + y

其中y为编号i书页对应范围内的书页。

那么，具体实现的话，需要求出每个i对应的y，这里用前缀和。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #define maxn 1000005
 5 #define lowbit(x) (-x&x)
 6 using namespace std;
 7 
 8 int n,high = 0;
 9 
10 struct node{
11     int pos,ai;
12 }list[maxn];
13 
14 int Tree[maxn],DP[maxn],minn;
15 int sum(int p){
16     int ans = 0;
17     while(p){
18 //        cout << ‘^‘;
19 //        cout << p;
20         ans += Tree[p];
21         p -= lowbit(p);
22     }
23     return ans;
24 }
25 void add(int p){
26     while(p <= high){
27 //        cout << ‘G‘;
28         Tree[p]++;
29         p += lowbit(p);
30 //        cout << p;
31     }
32 }
33 
34 bool cmp(const node &a,const node &b){
35     return a.pos < b.pos;
36 }
37 
38 int main(){
39     scanf("%d",&n);
40     
41     for(int i = 1;i <= n;i++){
42         scanf("%d%d",&list[i].pos,&list[i].ai);
43         list[i].pos++;
44         high = max(high,list[i].pos);
45     }
46     
47     sort(list+1,list+1+n,cmp);
48     
49 //    printf("#%d#\n",high);
50     
51     for(int i = 1;i <= n;i++){
52         
53 //        cout << ‘A‘;
54         add(list[i].pos);
55     }
56     
57     for(int i = 1;i <= n;i++){
58         int a = list[i].pos-1;
59         int b = list[i].pos-list[i].ai-1;
60         if(a < 0) a = 0;
61         if(b < 0) b = 0;
62         int y = sum(a)-sum(b);
63         if(i-y-1 < 0) continue;
64         DP[i] = DP[i-y-1] + y;
65     }
66     
67     minn = DP[n];
68     
69     for(int i = 1;i <= n;i++){
70         minn = min(minn,DP[i]+n-i);
71     }
72     
73     printf("%d",minn);
74     
75 //    cout << endl;
76 //    for(int i = 0;i <= n;i++) printf("%d ",DP[i]);
77     
78 //    cout << endl;
79 //    for(int i = 0;i <= 10;i++) printf("%d ",sum(i));
80     
81     return 0;
82 }

震惊！CYC的背景居然隐藏着这样的...