hdu4512

题意为求给定序列中单调递增子序列，且相邻的两个小标之差大于d；

用线段树可解，(单点更新，查询线段)  线段树中的点是存以a【i】结尾的所要求的字串的最大值，动态规划也可解，但自己不明白如何变形，路过大神留下意见；

线段树代码：

#include<stdio.h>#include<algorithm>#include<string.h>const int maxn=100110;int tree[4*maxn];int maxnum(int a,int b) {return a>b?a:b;}void pushup(int rt){    tree[rt]=maxnum(tree[rt<<1],tree[rt<<1|1]);}void buildtree(int l,int r,int rt){    if(l==r) { tree[rt]=0;return;}    int mid=(l+r)/2;    buildtree(l,mid,rt<<1);    buildtree(mid+1,r,rt<<1|1);    pushup(rt);}void update(int rt,int pos,int val,int left,int right){    if(left==right)    {        tree[rt]=val;        return ;    }    int mid=(left+right)/2;    if(pos<=mid) update(rt<<1,pos,val,left,mid);    else update(rt<<1|1,pos,val,mid+1,right);    pushup(rt);}int query(int l,int r,int rt,int left,int right){    if(l>r) return 0;    if(left==l&&right==r)        return tree[rt];    int mid=(left+right)/2;    if(l>mid) return query(l , r , rt<<1|1 ,mid+1,right );    else if(r<=mid) return query( l, r , rt<<1,left,mid );    else return maxnum(query(l, mid, rt<<1,left,mid),query( mid+1, r, rt<<1|1,mid+1,right));}struct ss{    int num;    int pos;}a[maxn];bool cmp(ss t1,ss t2){    if(t1.num==t2.num) return t1.pos>t2.pos;    else return t1.num<t2.num;}int main(){    //freopen("Input.txt","r",stdin);    int i,n,d;    while(~scanf("%d%d",&n,&d))    {        buildtree(1,n,1);        for(i=1;i<=n;i++)        {            scanf("%d",&a[i].num);            a[i].pos=i;        }        std::sort(a+1,a+n+1,cmp);        int ans=0;        for(i=1;i<=n;i++)        {            int pos=a[i].pos;            int tmp=query(1,pos-d-1,1,1,n);            ans=maxnum(ans,tmp+1);            update(1,pos,tmp+1,1,n);        }        printf("%d\n",ans);    }    return 0;}