UVa 1471 Defense Lines (二分+set优化)

题意：给定一个序列，然后让你删除一段连续的序列，使得剩下的序列中连续递增子序列最长。

析：如果暴力枚举那么时间复杂度肯定受不了，我们可以先进行预处理，f[i] 表示以 i 结尾的连续最长序列，g[i] 表示以 i 开头的连续最长序列，然后再去找最长的，

枚举 i，然后用set来维护一个单调上升的序列，我们把已经扫过的用set处理，按a[i]从小到大排序，然后f[i]也是从小到大，把不是最优的解全删掉，从而减少的要遍历的数目。

然后动态处理每个值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const LL mod = 1e3 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int g[maxn], f[maxn];
struct Node{
    int id, num;
    Node(int i, int n) : id(i), num(n) { }
    bool operator < (const Node &p) const{
        return id < p.id;
    }
};
set<Node> sets;
set<Node> :: iterator it;

int main(){
    int T;  cin >> T;
    while(T--){
        sets.clear();
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)  scanf("%d", a+i);
        f[0] = 1;
        for(int i = 1; i < n; ++i)
            if(a[i] > a[i-1])  f[i] = f[i-1] + 1;
            else f[i] = 1;
        g[n-1] = 1;
        for(int i = n-2; i >= 0; --i)
            if(a[i] < a[i+1])  g[i] = g[i+1] + 1;
            else g[i] = 1;
        int ans = f[0] + g[0] - 1;
        sets.insert(Node(a[0], 1));
        for(int i = 1; i < n; ++i){
            Node u(a[i], f[i]);
            it = sets.lower_bound(u);
            bool ok = true;
            if(it != sets.begin()){
                --it;
                ans = Max(ans, g[i] + it->num);
                if(it->num >= f[i]) ok = false;
            }
            if(ok){
                sets.erase(u);
                while(it != sets.end() && it->id > a[i] && it->num <= f[i])  sets.erase(it++);
                sets.insert(u);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}