題目鏈接

BZOJ2800

題解

區間加極難操作，差分之後可轉化為兩點一加一減
那麽現在問題就將每個點暫時獨立開來

先判定每個點是否被\((A,B)\)整除，否則無解
之後我們先將\(A,B\)化為互質，所有數除一個\((A,B)\)
求得
\[Ax + By = 1\]
那麽對於點\(d[i]\)，滿足
\[d[i] = A(xd[i] + kB) + B(yd[i] - kA)\]
其中\(k\)可以取任意值
我們對於單點的目標，是最小化
\[|xd[i] + kB|+|yd[i] - kA|\]
兩個絕對值相加是一個單峰函數，利用三分法即可得出\(k\)
從而得到每個點目前最優解\(X[i] = xd[i] + kB,Y[i] = yd[i] - kA\)

但是我們做到了單個點最優，但整體不一定合法，我們必須滿足正負操作次數相同
即
\[A\sum\limits_{i = 1}^{n}X[i] + B\sum\limits_{i = 1}^{n}Y[i] = 0\]
而由於\(\sum d[i] = 0\)
故我們只需保證\(T = \sum X[i] = 0\)
顯然我們只需改變\(\frac{T}{B}\)次
對於每個\(X[i]\)我們計算出它改變一次的代價，用一個堆維護即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
 

#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define cls(s) memset(s,0,sizeof(s))
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0
 
,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
struct pr{LL v,i;};
inline bool operator <(const pr& a,const pr& b){
    return a.v > b.v;
}
priority_queue<pr> q;
LL n,A,B,X,Y,d[maxn],h[maxn],xx[maxn],yy[maxn],dd;
void exgcd(LL a,LL b,LL& d,LL& x,LL& y){
    if (!b){d = a; x = 1; y = 0;}
    else exgcd(b,a % b,d,y,x),y -= (a / b) * x;
}
inline LL cost(int i,LL k){
    return abs(X * d[i] + k * B) + abs(Y * d[i] - k * A);
}
void workmin(){
    REP(i,n){
        LL l = -INF,r = INF,lmid,rmid,L,R,K;
        while (r - l >= 3){
            lmid = (l + l + r) / 3;
            rmid = (r + l + r) / 3;
            L = cost(i,lmid);
            R = cost(i,rmid);
            if (L == R){
                if (cost(i,lmid - 1) < L) r = rmid;
                else l = lmid;
            }
            else if (L > R) l = lmid;
            else r = rmid;
        }
        K = l;
        for (int j = l + 1; j <= r; j++)
            if (cost(i,j) < cost(i,K)) K = j;
        xx[i] = X * d[i] + K * B;
        yy[i] = Y * d[i] - K * A;
    }
}
inline LL price(int i){
    return abs(yy[i] - dd * A) + abs(xx[i] + dd * B) - abs(xx[i]) - abs(yy[i]);
}
void print(){
    //REP(i,n) printf("(%lld,%lld)\n",xx[i],yy[i]); puts("");
    LL ans = 0;
    REP(i,n) ans += abs(xx[i]) + abs(yy[i]);
    printf("%lld\n",ans >> 1);
}
void workok(){
    LL sum = 0;
    REP(i,n) sum += xx[i];
    sum /= B;
    dd = sum > 0 ? -1 : 1; sum = abs(sum);
    REP(i,n) q.push((pr){price(i),i});
    pr u;
    while (sum--){
        u = q.top(); q.pop();
        xx[u.i] += dd * B;
        yy[u.i] -= dd * A;
        q.push((pr){price(u.i),u.i});
    }
}
int main(){
    n = read(); A = read(); B = read(); LL D;
    exgcd(A,B,D,X,Y); A /= D; B /= D;
    REP(i,n){
        h[i] = read(); 
        if (h[i] % D){puts("-1"); return 0;}
        h[i] /= D; d[i] = h[i] - h[i - 1];
    }
    n++;
    d[n] = -h[n - 1];
    workmin();
    workok();
    print();
    return 0;
}

BZOJ2800 [Poi2012]Leveling Ground 【擴展歐幾裏得 + 三分 + 堆】