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HDU2227 Find the nondecreasing subsequences

题解:

dp+树状数组优化

dp[i]表示以i结尾时的非递减数组和.这样做的复杂度是o(n^2),不行

观察这个dp

for(int i=1;i<=n;i++){
    dp[i]=1;
    for(int j=1;j<i;j++)  if(a[j]<=a[i]) dp[i]=(dp[i]+dp[j])%mod;
}

dp[i]求的是前面全部小于等于a[i]的a[j]的dp[j]的和

那么,可以每次把dp[j]加入到树状数组中,

也就是每次 dp[i]=sum(a[i]);求的是每次小于等于a[i]的dp和,然后add(a[i],dp[i]+1)

最后求sum(maxn)即可

代码:

o(n^2) TLE

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<map>#include<set>using namespace std;using namespace std;#define pb push_back#define mp make_pair#define se second#define fs first#define ll long long#define CLR(x) memset(x,0,sizeof x)#define SZ(x) ((int)(x).size())#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)typedef pair<int,int> P;const double eps=1e-9;const int maxn=100010;const int mod=1000000007;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();}    while(ch>=0&&ch<=9){x=x*10+ch-0;ch=getchar();}    return x*f;}//-----------------------------------------------------------------------------int a[maxn];int dp[maxn];int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i]);    for(int i=1;i<=n;i++){        dp[i]=1;        for(int j=1;j<i;j++) if(a[j]<=a[i]) dp[i]=(dp[i]+dp[j])%mod;    }    int sum=0;    for(int i=1;i<=n;i++) sum=(sum+dp[i])%mod;    printf("%d\n",sum);}

o(nlog(n))

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<map>#include<set>using namespace std;using namespace std;#define pb push_back#define mp make_pair#define se second#define fs first#define ll long long#define CLR(x) memset(x,0,sizeof x)#define SZ(x) ((int)(x).size())#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)typedef pair<int,int> P;const double eps=1e-9;const int maxn=100010;const int mod=1000000007;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();}    while(ch>=0&&ch<=9){x=x*10+ch-0;ch=getchar();}    return x*f;}//-----------------------------------------------------------------------------pair<int,int> p[maxn];int b[maxn],c[maxn];int dp[maxn];int lowbit(int x) {return x&-x;}void add(int x,int v){    while(x<maxn){        c[x]+=v;        c[x]%=mod;        x+=lowbit(x);    }}int sum(int x){    int cnt=0;    while(x){        cnt+=c[x];        cnt%=mod;        x-=lowbit(x);    }    return cnt;}int main(){    int n;    while(~scanf("%d",&n)){    CLR(c);    for(int i=1;i<=n;i++){        scanf("%d",&p[i].fs);        p[i].se=i;        b[i]=p[i].fs;    }    sort(p+1,p+n+1);    for(int i=1;i<=n;i++) b[p[i].se]=i;    for(int i=1;i<=n;i++){        dp[i]=sum(b[i]);        add(b[i],dp[i]+1);    }    printf("%d\n",sum(maxn));    }}

 

 

HDU2227 Find the nondecreasing subsequences