hdu_1950_Bridging signals(LIS)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1950

题意：实际就是求最长递增子序列

题解：有两种解法，一种是利用二分，一种是用线段树

这个是这题的二分代码：

#include <cstdio>
#include<algorithm>
#define F(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N = 1e5+7;
int a
,d
;
int LIS(int* a, int n, int* d){
int len=1;d[1]=a[1];
F(i,2,n)if(d[len]<a[i])d[++len]=a[i];
else d[lower_bound(d+1,d+1+len,a[i])-d]=a[i];
return len;
}
int main(){
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i =1; i <=n; i++)scanf("%d",&a[i]);
printf("%d\n",LIS(a,n,d));
}
return 0;
}这个是求LIS的线段树的代码
#include<cstdio>
#include<algorithm>
#define root 1,n,1
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define F(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=1e5+7;
int n,sum[N<<2],a
,ans,dp
;

void update(int x,int k,int l,int r,int rt){
if(l==r){sum[rt]=k;return;}
int m=(l+r)>>1;
if(x<=m)update(x,k,ls);
else update(x,k,rs);
sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}

int query(int L,int R,int l,int r,int rt){
if(L<=l&&r<=R)return sum[rt];
int m=(l+r)>>1,ret=0;
if(L<=m)ret=max(ret,query(L,R,ls));
if(m<R)ret=max(ret,query(L,R,rs));
return ret;
}

int main(){
while(~scanf("%d",&n)){
F(i,1,n)scanf("%d",a+i),dp[i]=0;
F(i,0,(N<<2)-1)sum[i]=0;ans=0;
F(i,1,n){
if(a[i]-1>0){
dp[i]=max(dp[i],query(1,a[i]-1,root))+1;
update(a[i],dp[i],root);
}else dp[i]=1,update(a[i],dp[i],root);
ans=max(dp[i],ans);
}
printf("%d\n",ans);
}
return 0;
}