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CF1523G Try Booking 题解

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Solution

我好蠢啊。。。

考虑如果x确定怎么办?我们可以直接递归,每次处理 [l,r] 的最大贡献,然后找到在这个区间里面编号最小的区间,然后两边递归下去。考虑这样的复杂度,因为最多 nx 个区间,所以复杂度是 nxlog2n 的。

然后我们直接对于每个 x 做一遍就好了。3200 的清新题目。呜呜。。

Code

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#include <bits/stdc++.h> using namespace std; #define Int register int #define MAXN 100005 template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;} template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);} template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');} template <typename T> inline void chkmax (T &a,T b){a = max (a,b);} template <typename T> inline void chkmin (T &a,T b){a = min (a,b);} #define LOGN 205 #define ls(x) son[x][0] #define rs(x) son[x][1] int n,m,cnt,rt[MAXN],sl[MAXN],sr[MAXN],ind[MAXN],minv[MAXN * LOGN],son[MAXN * LOGN][2]; void modify (int &x,int l,int r,int pos,int miv){ if (!x) x = ++ cnt,minv[x] = 1e9; chkmin (minv[x],miv); if (l == r) return ; int mid = l + r >> 1; if (pos <= mid) modify (ls(x),l,mid,pos,miv); else modify (rs(x),mid + 1,r,pos,miv); } int query (int x,int l,int r,int ql,int qr){ if (!x) return 1e9; if (l >= ql && r <= qr) return minv[x]; int mid = l + r >> 1,res = 1e9; if (ql <= mid) chkmin (res,query (ls(x),l,mid,ql,qr)); if (qr > mid) chkmin (res,query (rs(x),mid + 1,r,ql,qr)); return res; } int lowbit (int x){return x & (-x);} void Modify (int l,int r,int p){ for (;r <= n;r += lowbit (r)) modify (rt[r],1,n,l,p); } int Query (int l,int r){ int res = 1e9; for (;r >= l;r -= lowbit (r)) chkmin (res,query (rt[r],1,n,l,n)); return res; } int solve (int l,int r){ if (l > r) return 0; int res = Query (l,r); if (res == 1e9) return 0; return sr[res] - sl[res] + 1 + solve (l,sl[res] - 1) + solve (sr[res] + 1,r); } int ans[MAXN]; signed main(){ read (n,m); for (Int i = 1;i <= m;++ i) read (sl[i],sr[i]),ind[i] = i; sort (ind + 1,ind + m + 1,[](int x,int y){return sr[x] - sl[x] > sr[y] - sl[y];}); for (Int x = n,i = 1;x >= 1;-- x){ while (i <= m && sr[ind[i]] - sl[ind[i]] + 1 == x) Modify (sl[ind[i]],sr[ind[i]],ind[i]),i ++; ans[x] = solve (1,n); } for (Int x = 1;x <= n;++ x) write (ans[x]),putchar ('\n'); return 0; }
posted @   Dark_Romance  阅读(41)  评论(2编辑  收藏  举报
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