LightOJ 1085(树状数组+离散化+DP，线段树)

All Possible Increasing Subsequences

Time Limit:3000MS      Memory Limit:65536KB      64bit IO Format:%lld & %llu

Appoint description: 

Description

An increasing subsequence from a sequence A₁, A₂ ... A_n is defined by A_i1, A_i2 ... A_ik, where the following properties hold

1. i₁ < i₂ < i₃ < ... < i_k and
2. 2.      A_i1 < A_i2 < A_i3 < ... < A_ik

Now you are given a sequence, you have to find the number of all possible increasing subsequences.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵) denoting the number of elements in the initial sequence. The next line will contain n integers separated by spaces, denoting the elements of the sequence. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case of input, print the case number and the number of possible increasing subsequences modulo 1000000007.

Sample Input

3

3

1 1 2

5

1 2 1000 1000 1001

3

1 10 11

Sample Output

Case 1: 5

Case 2: 23

Case 3: 7

题解：让求单调递增序列的个数；dp[i]=sum(dp[j])+1(j<i);由于数太大，需要离散化由于是单调的，所以当相等的时候树状数组要从大到小inset,以免对后面相等的造成影响，还可以用线段树写；

树状数组：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int MOD=1000000007;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=1e5+100;
int dp[MAXN],a[MAXN],p[MAXN];
int N;
int lowbit(int x){return x&(-x);}
void insert(int x,int v){
    while(x<=N){
        dp[x]=(dp[x]+v)%MOD;
        x+=lowbit(x);
    }
}
int sum(int x){
    int ans=0;
    while(x>0){
        ans=(ans+dp[x])%MOD;
        x-=lowbit(x);
    }
    return ans;
}
int cmp(int x,int y){
    if(a[x]!=a[y])return a[x]<a[y];
    else return x>y;
}
int main(){
    int T,kase=0;
    SI(T);
    while(T--){
        SI(N);
        for(int i=1;i<=N;i++)SI(a[i]),p[i]=i;
        sort(p+1,p+N+1,cmp);
        mem(dp,0);
        for(int i=1;i<=N;i++){
            insert(p[i],sum(p[i])+1);
        }
        printf("Case %d: %d\n",++kase,sum(N));
    }
    return 0;
}

线段树：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int MOD=1000000007;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int MAXN=1e5+100;
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define ll root<<1
#define rr root<<1|1
int tree[MAXN<<2],p[MAXN],a[MAXN];
int ans=0;
void pushup(int root){
    tree[root]=(tree[ll]+tree[rr])%MOD;
}
void insert(int root,int l,int r,int flog,int v){
    int mid=(l+r)>>1;
    if(l==flog&&r==flog){
        tree[root]=v;
        return;
    }
    if(mid>=flog)insert(lson,flog,v);
    if(mid<flog)insert(rson,flog,v);
    pushup(root);
}
void sum(int root,int l,int r,int L,int R){
    int mid=(l+r)>>1;
    if(l>=L&&r<=R){
        ans=(ans+tree[root])%MOD;
        return;
    }
    if(mid>=L)sum(lson,L,R);
    if(mid<R)sum(rson,L,R);
    return;
}
int cmp(int x,int y){
    if(a[x]!=a[y])return a[x]<a[y];
    else return x>y;
}
int main(){
    int T,N,kase=0;
    SI(T);
    while(T--){
        SI(N);
        for(int i=1;i<=N;i++)SI(a[i]),p[i]=i;
        sort(p+1,p+N+1,cmp);
        mem(tree,0);
        for(int i=1;i<=N;i++){
            ans=0;
            sum(1,1,N,1,p[i]);
            insert(1,1,N,p[i],ans+1);
        }
        printf("Case %d: %d\n",++kase,tree[1]);
    }
    return 0;
}