HDU 4604 Deque（dp、LIS）

原創

TaoSama

2020-02-21 10:27

題意:

給定N≤105的一個序列,現在有一個deque
將序列按順序往deque裏添加,只能向deque的頭或者尾添加,或者丟棄這個數
唯一的要求是deque裏的數必須是不降的,求deque裏的最多元素個數是多少

分析:

其實就是以Ax開頭的max(最長下降子序列與最長不降子序列的和,最長不升子序列與最長上升子序列的和)
題解的方法代碼很簡單
如果用BIT要搞2遍,數據水了一遍也過了,參考下列數據

2
6
4 5 4 6 7 8
6
4 3 4 6 7 8
ans:
6
6

代碼:

//
//  Created by TaoSama on 2016-02-26
//  Copyright (c) 2016 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 << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a[N];
int f[2][N], h[N];
//start from i, 0 for longest non-increasing, 1 for non-decreasing
int same[2][N]; //start from i, in LIS sequence, same as a[i]

void gao(int *f, int *same) {
    memset(h, 0x3f, sizeof h);
    for(int i = n; i; --i) {
        f[i] = upper_bound(h + 1, h + 1 + n, a[i]) - h;
        h[f[i]] = a[i];
        auto range = equal_range(h + 1, h + 1 + n, a[i]);
        same[i] = range.second - range.first;
    }
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)  scanf("%d", a + i);

        gao(f[0], same[0]);
        for(int i = 1; i <= n; ++i) a[i] = -a[i];
        gao(f[1], same[1]);

        int ans = 0;
        for(int i = 1; i <= n; ++i)
            ans = max(ans, f[0][i] + f[1][i] - min(same[0][i], same[1][i]));

        printf("%d\n", ans);
    }
    return 0;
}

給個搞一遍的代碼吧,懶得再寫再搞一遍了

代碼:

//
//  Created by TaoSama on 2016-02-26
//  Copyright (c) 2016 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 << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, a[N];

struct BIT {
    int n, b[N];
    void init(int _n) {
        n = _n;
        memset(b, 0, sizeof b);
    }
    void add(int i, int v, int delta) {
        for(; i && i <= n; i += delta * (i & -i))
            b[i] = max(b[i], v);
    }
    int sum(int i, int delta) {
        int ret = 0;
        for(; i && i <= n; i -= delta * (i & -i))
            ret = max(ret, b[i]);
        return ret;
    }
} bit;

int f[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        vector<int> xs;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            xs.push_back(a[i]);
        }
        sort(xs.begin(), xs.end());
        xs.resize(unique(xs.begin(), xs.end()) - xs.begin());

        //start from i, longest non-increasing, to left
        bit.init(xs.size());
        for(int i = n; i; --i) {
            a[i] = lower_bound(xs.begin(), xs.end(), a[i]) - xs.begin() + 1;
            f[i] = bit.sum(a[i], 1) + 1;
            bit.add(a[i], f[i], 1);
        }

        //start from i, longest non-decreasing, to right
        bit.init(xs.size());
        int ans = 0;
        for(int i = n; i; --i) {
            int cur = bit.sum(a[i], -1) + 1;
            bit.add(a[i], cur, -1);
            //ignore same numbers this time
            int rhs = bit.sum(a[i] + 1, -1);
            ans = max(ans, f[i] + rhs);
        }
        printf("%d\n", ans);
    }
    return 0;
}