老哥们,请问我做的对么?(记一次失败的st表乱搞)
今天a开始就不是很顺,然后到d,d努力读完题理解完题意,感觉自己又行了{
问最大的jump,我觉得如果单纯贪心策略显然会t,问min,max这类rmq果断上了st表(这东西我隔离的时候写的,没想到被拉出来用上了)。然后冥思苦想怎么才能满足跳的条件。
突然一个大胆的想法从脑中诞生,二分吧,自己动手模拟了下,感觉st表满足单调性的性质,毕竟min和max是区间内,只有扩大和缩小区间才能对其产生影响,而这种影响应该是符合单调性(然而没想到的一点是少考虑了一些问题)。然后复杂度一计算,nlogn刚好可以过!
怀着激动的心情码了出来,提交然后wa5,一直到最后也没调出来,我吐了,后来发现正解是单调栈,但二分st表并非不行(应该多一步就对了,去重是可以O(1)得到的
哎还是记录一下吧,好好练习,专心刷题,不要三心二意呢
#include <bits/stdc++.h> using namespace std; #define limit (500000 + 5)//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-6 #define FASTIO ios::sync_with_stdio(false);cin.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a,b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout) #define debug(x) cout<<x<<endl typedef long long ll; typedef unsigned long long ull; inline ll read(){ ll sign = 1, x = 0;char s = getchar(); while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();} while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();} return x * sign; }//快读 void write(ll x){ if(x < 0) putchar('-'),x = -x; if(x / 10) write(x / 10); putchar(x % 10 + '0'); } int n,m; int kase; int a[limit]; int f[limit][1<<6] , g[limit][1<<6]; void build(){ rep(i , 1, n){ f[i][0] = g[i][0] = a[i]; } for(int j = 1; (1<<j) <= n; ++j){ for(int i = 1; i + (1<<j) - 1 <= n; ++i){ f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]); g[i][j] = min(g[i][j - 1], g[i + (1 << (j - 1))][j - 1]); } } } int query(int l, int r){ int tmp = log2(r - l + 1); return min(g[l][tmp] , g[r - (1 << tmp) + 1][tmp]); } int querymax(int l, int r){ int tmp = log2(r - l + 1); return max(f[l][tmp] , f[r - (1 << tmp) + 1][tmp]); } bool judge(int st, int x){//开始和x bool flag = querymax(st + 1, st + x) < min({a[st], a[st + x + 1]}); bool flag2 = query(st + 1, st + x) > max(a[st], a[st + x + 1]); return flag || flag2; } int check(int st){ int l = 1, r = n - st; int mid; while (l <= r){ mid = l + (r - l) / 2; if(judge(st,mid))l = mid + 1; else r = mid - 1; } return l + (r - l) / 2; } int main() { #ifdef LOCAL FOPEN; #endif cin>>n; rep(i ,1,n){ a[i] = read(); } build(); ll ans = 0; for(int i = 1 ; i < n ; i += check(i),++ans); cout<<ans<<endl; return 0; }
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