CF 462 C. A Twisty Movement 分段想 线段树 或 dp
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题意
有一个只包含1和2的序列,试翻转一个区间,使得结果中非连续非递减数列最长。
思路
一、
作出1的前缀计数和为cnt1,2的后缀计数和为cnt2, 由于要找出【1,1,1】【2,2,2】【1,1,1】【2,2,2】的四段,设中间的分割点是p,k,q,可得到
ans=cnt1[p]+cnt2[p+1]−cnt2[k+1]+cnt1[q]−cnt1[k]+cnt2[q+1]ans=cnt1[p]+cnt2[p+1]−cnt2[k+1]+cnt1[q]−cnt1[k]+cnt2[q+1]
化简得到
ans=cnt1[p]+cnt2[p+1]+cnt1[q]+cnt2[q+1]−cnt1[k]−cnt2[k+1]ans=cnt1[p]+cnt2[p+1]+cnt1[q]+cnt2[q+1]−cnt1[k]−cnt2[k+1]
所以我们可以枚举k, 用线段树维护cnt1【i】 + cnt2【i+1】 的最大值,我这样的公式,需要线段树区间为【0,n】。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ \\ Λ_Λ 来了老弟 \('ㅅ') > ⌒ヽ / へ\ / / \\ レ ノ ヽ_つ / / / /| ( (ヽ | |、\ | 丿 \ ⌒) | | ) / 'ノ ) Lノ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 2009; int a[maxn],cnt1[maxn],cnt2[maxn]; int t[maxn<<2]; void build(int l,int r,int rt){ if(l == r){ t[rt] = cnt1[l] + cnt2[l+1]; return ; } int mid = (l + r) >> 1; build(l, mid, rt<<1); build(mid+1,r,rt<<1|1); t[rt] = max(t[rt<<1], t[rt<<1|1]); } int query(int L,int R,int l,int r,int rt){ if(l >= L && r <= R){ return t[rt]; } int mx = 0; int mid = (l + r) >> 1; if(mid >= L) mx = max(mx, query(L, R, l, mid, rt<<1)); if(mid < R) mx = max(mx, query(L, R, mid+1, r, rt<<1|1)); return mx; } int main(){ int n; scanf("%d", &n); rep(i, 1, n) scanf("%d", &a[i]); rep(i, 1, n){ if(a[i] == 1) cnt1[i] = cnt1[i-1] + 1; else cnt1[i] = cnt1[i-1]; } for(int i=n; i>=1; i--) { if(a[i] == 2) cnt2[i] = cnt2[i+1] + 1; else cnt2[i] = cnt2[i+1]; } build(0, n, 1); int ans = 1; for(int i=1; i<=n; i++){ int tmp = query(0,i ,0,n,1) + query(i,n,0,n,1) - cnt1[i] - cnt2[i+1]; ans = max(ans, tmp); } cout<<ans<<endl; return 0; }View Code
二、
这道题中分四段是关键,我们可以定义dp【i】【1】表示1~i个分一段的答案,dp【i】【2】表示分两段,dp【i】【3】表示分三段,dp【i】【4】表示分四段。
/*-----------------------showtime----------------------*/ int dp[5]; int main(){ int n; scanf("%d", &n); rep(i, 1, n) { int x; scanf("%d", &x); dp[1] = dp[1] + (x == 1); dp[2] = max(dp[1], dp[2] + (x==2)); dp[3] = max(dp[2], dp[3] + (x==1)); dp[4] = max(dp[3], dp[4] + (x==2)); } cout<<dp[4]<<endl; }View Code
标签:Twisty,int,CF,462,cnt2,cnt1,include,dp,define 来源: https://www.cnblogs.com/ckxkexing/p/10367409.html