NEERC 17 G.The Great Wall

Solution

這題的第一步$trick$是：我們注意到取$a,b,c$的集合兩兩不交且并集為$U$，因此確定了$b,c$之后可以簡單地唯一確定$a$，于是我們通過讓$b_i?=a_i,c_i?=a_i,Ans+=\sum a_i$來削減一維。

然后令$b_i^{\prime }={\sum }_{j=1}^i{b}_i$，$c_i^{\prime }={\sum }_{j=1}^i{c}_i$，二分答案$k$。

• 當$x+r?1\ge y$，$sum=c_{x+r?1}^{\prime }?c_{y?1}^{\prime }+b_{y?1}^{\prime }?b_{x?1}^{\prime }+b_{y+r?1}^{\prime }?b_{x+r?1}^{\prime }=(c_{x+r?1}^{\prime }?b_{x?1}^{\prime }?b_{x+r?1}^{\prime })+(?c_{y?1}^{\prime }+b_{y?1}^{\prime }+b_{y+r?1}^{\prime })\le k$，可以簡單地用二維數(shù)點解決。
• 當$x+r?1<y$，$sum=(b_{x+r?1}^{\prime }?b_{x?1}^{\prime })+(b_{y+r?1}^{\prime }?b_{y?1}^{\prime })\le k$，同樣可以簡單地用二維數(shù)點解決。

時間復(fù)雜度$O(nl{g}^{2}n)$。

Code

#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <ctime> #include <cassert> #include <string.h> //#include <unordered_set> //#include <unordered_map> //#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; } template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int,int> PR; typedef vector<int> VI;const lod eps=1e-11; const lod pi=acos(-1); const int oo=1<<30; const ll loo=1ll<<62; const int mods=998244353; const int MAXN=600005; const int INF=0x3f3f3f3f;//1061109567 /*--------------------------------------------------------------------*/ inline ll read() {ll f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f; } int n,r,s[MAXN]; ll p[MAXN],b[MAXN],c[MAXN],B[MAXN],a[MAXN],k; void add(int x,int y) { if (!x) return; for (;x<=n-r+1;x+=x&(-x)) s[x]+=y; } int query(int x) { upmin(x,n-r+1); int ans=0; for (;x;x-=x&(-x)) ans+=s[x]; return ans; } ll solve1(ll x) {ll ans=0;for (int i=1;i<=n-r+1;i++) B[i]=p[i]=-c[i-1]+b[i-1]+b[i+r-1];sort(p+1,p+n-r+2);for (int i=1;i<=n-r+1;i++) B[i]=lower_bound(p+1,p+n-r+2,B[i])-p;for (int i=1;i<=r-1;i++) add(B[i],1);for (int i=1;i<=n-r+1;i++){add(B[i],-1);if (i<=n-r-r+2) add(B[i+r-1],1);ans+=query(upper_bound(p+1,p+n-r+2,x-(c[i+r-1]-b[i-1]-b[i+r-1]))-p-1);}return ans; } ll solve2(ll x) {ll ans=0;for (int i=1;i<=n-r+1;i++) B[i]=p[i]=b[i+r-1]-b[i-1];sort(p+1,p+n-r+2);for (int i=1;i<=n-r+1;i++) B[i]=lower_bound(p+1,p+n-r+2,B[i])-p;for (int i=r;i<=n-r+1;i++) add(B[i],1);for (int i=1;i<=n-r+1;i++){if (i<=n-r-r+2) add(B[i+r-1],-1);ans+=query(upper_bound(p+1,p+n-r+2,x-(b[i+r-1]-b[i-1]))-p-1);}return ans; } signed main() {ll sum=0;n=read(),r=read(),k=read();for (int i=1;i<=n;i++) a[i]=read(),sum+=a[i];for (int i=1;i<=n;i++) b[i]=read()-a[i]+b[i-1];for (int i=1;i<=n;i++) c[i]=read()-a[i]+c[i-1];ll l=0,r=1000000ll*n;while (l<r){ll mid=(l+r)>>1,t=solve1(mid)+solve2(mid);if (t>=k) r=mid;else l=mid+1;}printf("%lld\n",r+sum);return 0; }

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