UVA10534-Wavio Sequence（LIS）

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题目链接

http://acm.hust.edu.cn/vjudge/problem/19219

思路

只需正向和逆向分别做一次LIS，然后ans = 2 * min(d1[i], d2[i]) - 1

代码

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endl#define CLR(a, x) memset(a, x, sizeof(a))#define FILL(a, n, x) for (int i = 0; i < n; i++) a[i] = xconst int maxn = 10000 + 5;int n, d1[maxn], d2[maxn], a[maxn], g1[maxn], g2[maxn];int main() {    while (~scan(n)) {        for (int i = 0; i < n; i++) scan(a[i]);        for (int i = 1; i <= n; i++) g1[i] = INT_MAX;        for (int i = 0; i < n; i++) {            int k = lower_bound(g1 + 1, g1 + 1 + n, a[i]) - g1;            d1[i] = k;            g1[k] = a[i];        }        for (int i = 1; i <= n; i++) g2[i] = INT_MAX;        for (int i = n - 1; i >= 0; i--) {            int k = lower_bound(g2 + 1, g2 + 1 + n, a[i]) - g2;            d2[i] = k;            g2[k] = a[i];        }        int res = 1;        for (int i = 0; i < n; i++) {            int t = 2 * min(d1[i], d2[i]) - 1;            res = max(res, t);        }        printf("%d\n", res);    }    return 0;}

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