2017北理校赛H题 青蛙过河（线段树, dp, 离散化）

题面：

这题的O(n^2) dp很容易想出来，由于我没有数据，也没法提交测试。所以就拿这个dp来简单对拍了。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1010;
int dp[maxn];
int n, c;
int a[maxn];

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d%d",&n,&c)) {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        dp[1] = 1;
        for(int i = 2; i <= n; i++) {
            dp[i] = 1;
            for(int j = 1; j < i; j++) {
                if((int)abs(a[i]-a[j]) >= c) dp[i] = max(dp[i], dp[j]+1);
            }
        }
        printf("%d
", dp[n]);
    }
    return 0;
}

这道题的n非常大，显然这个O(n^2)的dp是非常不给力的，考虑转移状态：dp(i)一定由之前的dp(j)转移过来，并且这个dp(j)一定是a(j)符合条件并且dp(j)是最大的。可以用线段树维护某个数字x的[-inf,x-c]与[x+c,inf]的dp值，这样查询的时候可以直接查询这两个区间，更新的时候只需要更新x点的dp值就行了。

数据范围很大，有1e9，因此要离散化。离散化也很好操作，考虑查询，每一个点有2个查询的点：x-c和x+c，加上-inf和inf再加上x，把这几个部分离散化就行了。

#include <bits/stdc++.h>
using namespace std;

#define lrt rt << 1
#define rrt rt << 1 | 1
typedef long long LL;
typedef struct Seg {
    LL sign, val;
}Seg;
const LL maxn = 300300;
const LL inf = 2000000000LL;
Seg seg[maxn<<3];
LL n, c;
LL a[maxn], h[maxn*4], hcnt;
LL id(LL x) { return lower_bound(h, h+hcnt, x) - h + 1; }
void pushup(LL rt) { seg[rt].val = max(seg[lrt].val, seg[rrt].val); }

void build(LL l, LL r, LL rt) {
    seg[rt].sign = -1; seg[rt].val = 0;
    if(l == r) return;
    LL mid = (l + r) >> 1;
    build(l, mid, lrt);
    build(mid+1, r, rrt);
}

void pushdown(LL rt) {
    if(~seg[rt].sign) {
        seg[lrt].sign = seg[rrt].sign = seg[rt].sign;
        seg[lrt].val = seg[rrt].val = seg[rt].sign;
        seg[rt].sign = -1;
    }
}

void update(LL pos, LL val, LL l, LL r, LL rt) {
    if(l == r) {
        seg[rt].val = seg[rt].sign = val;
        return;
    }
    pushdown(rt);
    LL mid = (l + r) >> 1;
    if(pos <= mid) update(pos, val, l, mid, lrt);
    else update(pos, val, mid+1, r, rrt);
    pushup(rt);
}

LL query(LL L, LL R, LL l, LL r, LL rt) {
    if(L <= l && r <= R) return seg[rt].val;
    pushdown(rt);
    LL mid = (l + r) >> 1;
    LL ret = -1;
    if(L <= mid) ret = max(ret, query(L, R, l, mid, lrt));
    if(mid < R) ret = max(ret, query(L, R, mid+1, r, rrt));
    return ret;
}

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%lld%lld",&n,&c)) {
        hcnt = 0; h[hcnt++] = -inf; h[hcnt++] = inf;
        for(LL i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            h[hcnt++] = a[i] - c;
            h[hcnt++] = a[i];
            h[hcnt++] = a[i] + c;
        }
        sort(h, h+hcnt); hcnt = unique(h, h+hcnt) - h;
        build(1, hcnt, 1);
        LL L = id(-inf), R = id(inf);
        LL dp = -1;
        for(LL i = 1; i <= n; i++) {
            LL l = id(a[i]-c), r = id(a[i]+c);
            dp = max(query(L, l, 1, hcnt, 1), query(r, R, 1, hcnt, 1)) + 1;
            update(id(a[i]), dp, 1, hcnt, 1);
        }
        printf("%lld
", dp);
    }
    return 0;
}