HDU 2227 Find the nondecreasing subsequences （线段树）
Find the nondecreasing subsequences
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1235 Accepted Submission(s): 431
Problem 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
Author
8600
Recommend
lcy
解题思路： 每进来一个数，方法数是原来所有小于等于它的数的方法数之和+1，然后再把这个数添加进为原来的数，下面依次循环。
100000个数，但是数值比较大，所以离散化一下，线段树来维护，每次lg的操作
因此总效率 O（n^lgn）
#include <iostream>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
const int maxn=100010;
const int yu= 1000000007;
int n;
unsigned int data[maxn];
map <unsigned int,int> mp;
struct node{
int l,r;
unsigned c;
}a[maxn*4];
void build(int l,int r,int k){
a[k].l=l;a[k].r=r;a[k].c=0;
if(l<r){
int mid=(l+r)/2;
build(l,mid,2*k);
build(mid+1,r,2*k+1);
}
}
void insert(int l,int r,int c,int k){
a[k].c=(a[k].c+c)%yu;
if(l<=a[k].l && a[k].r<=r) return;
else{
int mid=(a[k].l+a[k].r)/2;
if(l>=mid+1) insert(l,r,c,2*k+1);
else if(r<=mid) insert(l,r,c,2*k);
else{
insert(l,mid,c,2*k);
insert(mid+1,r,c,2*k+1);
}
}
}
unsigned query(int l,int r,int k){
if(l<=a[k].l && a[k].r<=r){
return a[k].c;
}else{
int mid=(a[k].l+a[k].r)/2;
if(l>=mid+1) return query(l,r,2*k+1);
else if(r<=mid) return query(l,r,2*k);
else{
return (query(l,mid,2*k)+query(mid+1,r,2*k+1))%yu;
}
}
}
void initial(){
mp.clear();
build(1,n,1);
}
void input(){
int cnt=1;
map <unsigned int,int>::iterator it;
for(int i=1;i<=n;i++){
scanf("%d",&data[i]);
mp[data[i]]=0;
}
for(it=mp.begin();it!=mp.end();it++){
it->second=cnt++;
}
}
void computing(){
int ans=0,tmp,pos;
for(int i=1;i<=n;i++){
pos=mp[data[i]];
tmp=query(1,pos,1)+1;
ans=(ans+tmp)%yu;
insert(pos,pos,tmp,1);
}
cout<<ans<<endl;
}
int main(){
while(scanf("%d",&n)!=EOF){
initial();
input();
computing();
}
return 0;
}