hdu_1950_Bridging signals(LIS)

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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[N],d[N]; 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[N],ans,dp[N];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;}

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