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:

 1 #include <cstdio>  
 2 #include<algorithm>
 3 #define F(i,a,b) for(int i=a;i<=b;i++)
 4 using namespace std;  
 5 const int N = 1e5+7;  
 6 int a[N],d[N]; 
 7 int LIS(int* a, int n, int* d){
 8     int len=1;d[1]=a[1];
 9     F(i,2,n)if(d[len]<a[i])d[++len]=a[i]; 
10     else d[lower_bound(d+1,d+1+len,a[i])-d]=a[i];
11     return len;  
12 }
13 int main(){
14     int t,n;
15     scanf("%d",&t);
16     while(t--){
17     scanf("%d",&n);
18         for(int i =1; i <=n; i++)scanf("%d",&a[i]); 
19         printf("%d\n",LIS(a,n,d));  
20     }
21     return 0;  
22 }
23 这个是求LIS的线段树的代码
24 
25 #include<cstdio>
26 #include<algorithm>
27 #define root 1,n,1
28 #define ls l,m,rt<<1
29 #define rs m+1,r,rt<<1|1
30 #define F(i,a,b) for(int i=a;i<=b;i++)
31 using namespace std;
32 const int N=1e5+7;
33 int n,sum[N<<2],a[N],ans,dp[N];
34 
35 void update(int x,int k,int l,int r,int rt){
36     if(l==r){sum[rt]=k;return;}
37     int m=(l+r)>>1;
38     if(x<=m)update(x,k,ls);
39     else update(x,k,rs);
40     sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
41 }
42 
43 int query(int L,int R,int l,int r,int rt){
44     if(L<=l&&r<=R)return sum[rt];
45     int m=(l+r)>>1,ret=0;
46     if(L<=m)ret=max(ret,query(L,R,ls));
47     if(m<R)ret=max(ret,query(L,R,rs));
48     return ret;
49 }
50 
51 int main(){
52     while(~scanf("%d",&n)){
53         F(i,1,n)scanf("%d",a+i),dp[i]=0;
54         F(i,0,(N<<2)-1)sum[i]=0;ans=0;
55         F(i,1,n){
56             if(a[i]-1>0){
57                 dp[i]=max(dp[i],query(1,a[i]-1,root))+1;
58                 update(a[i],dp[i],root);
59             }else dp[i]=1,update(a[i],dp[i],root);
60             ans=max(dp[i],ans);
61         }
62         printf("%d\n",ans);
63     }
64     return 0;
65 }
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posted @ 2016-06-30 12:02  bin_gege  阅读(112)  评论(0编辑  收藏  举报