HDU 5489 Removed Interval （dp+线段树） 2015合肥网络赛

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题意:

给定一个长度为N<=105的序列求删去长度为0<=L<=N的序列后的LIS

分析:

假设你已经会O(nlogn)的LIS, 删去i前面连续一段长度为L即[i−L,i−1]的序列, 左边是i−L−1, 右边是i, 求剩下的LIS
首先预处理出:
f[i]:=以a[i]结尾的LIS的最大长度
g[i]:=以a[i]开头的LIS的最大长度
我们可以根据i, 找到 [0,i−L−1]之间的一个值
其值小于a[i], 而其f[i]值是最大的, 也就是O(n2)求LIS的思想
关键是如何快速确定在[0,i−L−1]中, 找到这个值
我们利用线段树和离散化, 将a[i]映射到线段树上
假设i位置对应的值是x,并且其映射到线段树上对应的下标是y
那么我们只要在线段树上查找[0,y−1]之间的最大值maxv
maxv[i]=max{f[j],j∈[0,i−L+1]∪a[j]<a[i]}
其实就是dp[i]:=以a[i]结尾, 删去[i−L,i−1]长度为L的LIS的最大长度
dp[i]=maxv[i]+g[i],ans=max{dp[i]}

代码:

////  Created by TaoSama on 2015-09-27//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, L, a[N], all[N];int f[N], g[N], h[N];int maxv[N << 2];void update(int o, int v, int l, int r, int rt) {    if(l == r) {        maxv[rt] = max(maxv[rt], v);        return;    }    int m = l + r >> 1;    if(o <= m) update(o, v, l, m, rt << 1);    else update(o, v, m + 1, r, rt << 1 | 1);    maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]);}int query(int L, int R, int l, int r, int rt) {    if(L <= l && r <= R) return maxv[rt];    int m = l + r >> 1, ret = -INF;    if(L <= m) ret = max(ret, query(L, R, l, m, rt << 1));    if(R > m) ret = max(ret, query(L, R, m + 1, r, rt << 1 | 1));    return ret;}int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d%d", &n, &L);        for(int i = 1; i <= n; ++i) {            scanf("%d", a + i);            all[i] = a[i];        }        sort(all + 1, all + 1 + n);        int m = unique(all + 1, all + 1 + n) - all - 1;        memset(h, 0x3f, sizeof h);        for(int i = 1; i <= n; ++i) {            int k = lower_bound(h + 1, h + 1 + n, a[i]) - h;            f[i] = k;            h[k] = a[i];        }        memset(h, 0x3f, sizeof h);        for(int i = n; i; --i) {            int k = lower_bound(h + 1, h + 1 + n, -a[i]) - h;            g[i] = k;            h[k] = -a[i];        }        int ans = 0;        a[n + 1] = INF; g[n + 1] = 0;        memset(maxv, 0, sizeof maxv);        //delete [i-l,i-1] length:l   [0,i-l-1]->maxv        for(int i = L + 1; i <= n + 1; ++i) {            int o = lower_bound(all + 1, all + 1 + m, a[i]) - all;            ans = max(ans, query(0, o - 1, 0, m, 1) + g[i]);            o = lower_bound(all + 1, all + 1 + m, a[i - L]) - all; //插入一个新的            if(i <= n) update(o, f[i - L], 0, m, 1);        }        printf("Case #%d: %d\n", ++kase, ans);    }    return 0;}

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