Codeforces 474E Pillars dp+线段树（水

题目链接：点击打开链接

给定n d

下面n座山的高度。

任选一座山，每次可以跳到右边的任意距离的山，问最多能经过几座山，并输出路径。

且→跳时，两座山的高度差必须>=d

思路：

线段树维护高度为x的山的值和前驱。

高度离散一下。

然后这么搞搞。。

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <map>#include <vector>typedef long long ll;template <class T>inline bool rd(T &ret) {    char c; int sgn;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    ret*=sgn;    return 1;}template <class T>inline void pt(T x) {    if (x <0) {        putchar('-');        x = -x;    }    if(x>9) pt(x/10);    putchar(x%10+'0');}const int N = 300010;using namespace std;typedef pair<int,int> pii;#define L(x) tree[x].l#define R(x) tree[x].r#define Lson(x) (x<<1)#define Rson(x) (x<<1|1)#define P(x) tree[x].pstruct node{    int l, r;    pii p;}tree[N<<2];map<ll, int> mp;vector<ll>G;void push_up(int id){    P(id) = max(P(Lson(id)), P(Rson(id)));}void build(int l, int r, int id){    L(id) = l; R(id) = r;    P(id) = pii(0, -1);    if(l == r)        return ;    int mid = (l+r)>>1;    build(l, mid, Lson(id)); build(mid+1, r, Rson(id));}pii query(int l, int r, int id){    if( l == L(id) && R(id) == r)        return P(id);    int mid = (L(id) + R(id))>>1;    if(mid < l)        return query(l, r, Rson(id));    else if(r <= mid)        return query(l, r, Lson(id));    else        return max(query(l, mid, Lson(id)), query(mid+1, r, Rson(id)) );}void updata(int pos, pii p, int id){    if(L(id) == R(id))    {        P(id) = max(P(id), p);        return ;    }    int mid = (L(id) + R(id))>>1;    if(mid < pos)        updata(pos, p, Rson(id));    else if(pos <= mid)        updata(pos, p, Lson(id));    push_up(id);}ll a[N], d;int n, dp[N];void input(){    mp.clear();    G.clear();    for(int i = 1; i <= n; i++) {        rd(a[i]);        G.push_back(a[i]);        G.push_back(a[i]+d);        G.push_back(a[i]-d);    }    sort(G.begin(), G.end());    G.erase(unique(G.begin(), G.end()), G.end());    for(int i = 0; i < G.size(); i++) mp[G[i]] = i+1;}vector<int>hehe;int main(){    pii ans;    while(cin>>n>>d){        input();        build(1, (int)G.size(), 1);        for(int i = 1; i <= n; i++)        {            ans = max( query(1, mp[a[i]-d], 1), query(mp[a[i]+d], (int)G.size(), 1));            dp[i] = ans.second;            ans.first++; ans.second = i;            updata(mp[a[i]], ans, 1);        }        ans = query(1, (int)G.size(), 1);        int u = ans.second;        hehe.clear();        while(~u)            hehe.push_back(u), u = dp[u];        pt(hehe.size()); putchar('\n');        for(int i = hehe.size()-1; i >= 0; i--)            printf("%d%c", hehe[i], i==0?'\n':' ');    }    return 0;}/*6 34 1 3 0 3 0*/