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  • UVa 10534 Wavio Sequence (LIS+暴力)

    题意:给定一个序列,求一个最长子序列,使得序列长度为奇数,并且前一半严格递增,后一半严格递减。

    析:先正向和逆向分别求一次LIS,然后再枚举中间的那个数,找得最长的那个序列。

    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>
    #include <unordered_map>
    #include <unordered_set>
    #define debug() puts("++++");
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 10000 + 5;
    const int mod = 2000;
    const int dr[] = {-1, 1, 0, 0};
    const int dc[] = {0, 0, 1, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    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 bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    int dp1[maxn], dp2[maxn], dp[maxn];
    int a[maxn];
    
    int solve(){
        int ans = 0;
        for(int i = 1; i < n; ++i){
            ans = max(ans, min(dp1[i], dp2[i])*2 + 1);
        }
        return ans;
    }
    
    int main(){
        while(scanf("%d", &n) == 1){
            for(int i = 0; i < n; ++i)  scanf("%d", a+i);
            memset(dp, INF, sizeof dp);
            for(int i = 0; i < n; ++i){
                *lower_bound(dp, dp+n, a[i]) = a[i];
                dp1[i] = lower_bound(dp, dp+i+1, INF) - dp - 1;
            }
            memset(dp, INF, sizeof dp);
            for(int i = n-1; i >= 0; --i){
                *lower_bound(dp, dp+n, a[i]) = a[i];
                dp2[i] = lower_bound(dp, dp+n, INF) - dp - 1;
            }
            printf("%d
    ", solve());
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6558789.html
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