• HDU


    HDU - 5406

    排序之后就变成了求两个不相交的最长LIS, 用树状数组把n ^ 3 的dp 优化成 n ^ 2 * log(n) 就ok 了。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 1000 + 2;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}
    
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    
    int n, tot, dp[N][N], hs[N];
    
    struct Bit {
        int a[N];
        void init() {
            for(int i = 1; i <= tot; i++) {
                a[i] = 0;
            }
        }
        void modify(int x, int v) {
            for(int i = x; i <= tot; i += i & -i) {
                chkmax(a[i], v);
            }
        }
        int preMax(int x) {
            int ans = 0;
            for(int i = x; i; i -= i & -i) {
                chkmax(ans, a[i]);
            }
            return ans;
        }
    } big[N], sml[N];
    
    PII apple[N];
    int a[N];
    
    int main() {
    
        int T; scanf("%d", &T);
        while(T--) {
            scanf("%d", &n);
            tot = 0;
            for(int i = 1; i <= n; i++) {
                scanf("%d%d", &apple[i].fi, &apple[i].se);
                hs[++tot] = apple[i].se;
            }
            hs[++tot] = 0;
    
            if(n == 1) {
                puts("1");
                continue;
            }
            sort(apple + 1, apple + 1 + n, [&](PII a, PII b) {
                if(a.fi == b.fi) return a.se < b.se;
                else return a.fi > b.fi;
            });
            sort(hs + 1, hs + 1 + tot);
            tot = unique(hs + 1, hs + 1 + tot) - hs - 1;
    
            for(int i = 0; i <= n; i++) {
                a[i] = lower_bound(hs + 1, hs + 1 + tot, apple[i].se) - hs;
            }
    
            for(int i = 0; i <= n; i++) {
                big[i].init();
                sml[i].init();
            }
    
            int ans = 0;
            for(int i = 1; i <= n; i++) {
                for(int j = 0; j < i; j++) {
                    dp[i][j] = j ? 2 : 1;
                    chkmax(dp[i][j], sml[j].preMax(a[i]) + 1);
                    chkmax(dp[i][j], big[j].preMax(a[i]) + 1);
                    big[i].modify(a[j], dp[i][j]);
                    sml[j].modify(a[i], dp[i][j]);
                    chkmax(ans, dp[i][j]);
                }
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/11157277.html
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