CodeForces 631E Product Sum（斜率优化DP+二分|三分） ★

来源：互联网发布：大数据就业方向编辑：程序博客网时间：2024/05/21 13:30

题意：给出n个数，现在可以移动一个数的位置，现在要使和sigma（ai*i）最大，询问这个最大和。

思路：将一个数向左移动和向右移动是一样的，现在考虑向左移动。

先预处理出前缀和，将一个数向左移动后，那么改变量为sum[r-1]-sum[l-1]+a[r]*(r-l),考虑枚举r，那么和r有关的数据就变成了常量。

现在问题转化成了求a[r]*l-sum[l-1]，注意到这里l和sum[l-]都是递增的，所以可以考虑用斜率优化来加速dp，

维护一个下凸曲线，然后对于每一个a[r]，二分斜率或者三分截距就可以解决当前位置左移的最大值。

直接三分：

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>#include <cmath>#include <vector>#include <set>#include <list>#include <queue>#include <map>#include <bitset>using namespace std;#define L(i) i<<1#define R(i) i<<1|1#define INF  0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-3#define maxn 200100#define MOD 1000000007int n;long long a[maxn],cur;long long total,sum[maxn];long long solve(int x,int y){    if(x <= y)        return total - sum[y] + sum[x] + a[x] * (y - x);    return total + sum[x-1] - sum[y-1] + a[x] * (y - x);}int main(){    int t;    //scanf("%d",&t);    while(scanf("%d",&n) != EOF)    {        sum[0] = 0;        for(int i = 1; i <= n; i++)        {            scanf("%lld",&a[i]);            sum[i] = sum[i-1] + a[i];            total += i * a[i];        }        long long ans = -0x3f3f3f3f3f3f3f3f;        for(int i = 1; i <= n; i++)        {            int l = 1,r = n;            long long cur = -0x3f3f3f3f3f3f3f3f;            while(l + 1 <= r)            {                int mid = l + (r - l) / 3;                int mmid = r - (r - l) / 3;                long long ans1 = solve(i,mid);                long long ans2 = solve(i,mmid);                if(ans1 > ans2)                {                    r = mmid - 1;                    cur = max(cur,ans1);                }                else                {                    l = mid + 1;                    cur = max(cur,ans2);                }            }            cur = max(cur,solve(i,l));            ans = max(ans,cur);        }        printf("%lld\n",ans);    }    return 0;}

斜率优化：

#include<bits/stdc++.h>#define eps 1e-6#define LL long long#define pii pair<int, int>#define pb push_back#define mp make_pair//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 200020;const LL INF = 1e17;int n;int a[MAXN];LL sumv[MAXN];int Q[MAXN], h, t;struct Point {LL x, y;Point(LL _x = 0, LL _y = 0) : x(_x), y(_y) {}} p[MAXN];LL getY(int k, int j) {return p[k].y-p[j].y;}LL getX(int k, int j) {return p[k].x-p[j].x;}bool check1(int k, int j) {return getY(Q[k+1], Q[k]) >= getX(Q[k+1], Q[k])*a[j]; }bool check2(int k, int j) {return getY(Q[k], Q[k+1]) <= getX(Q[k], Q[k+1])*a[j];}int main(){    //freopen("input.txt", "r", stdin);scanf("%d", &n);LL ans = 0;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);sumv[i] = sumv[i-1] + a[i];ans += (LL)i * a[i];}for (int i = 1; i <= n; i++) p[i] = Point(i, sumv[i-1]);LL tmp = ans;t = 0;for (int i = 2; i <= n; i++) {while (t > 1 && getY(i-1, Q[t])*getX(Q[t], Q[t-1]) <= getY(Q[t], Q[t-1])*getX(i-1, Q[t]))t--;Q[++t] = i-1;int l = 1, r = t;while (l < r) {int mid = (l+r) >> 1;if (check1(mid, i)) r = mid;else l = mid + 1;} ans = max(ans, tmp+(LL)a[i]*Q[r]-sumv[Q[r]-1]+sumv[i-1]-(LL)i*a[i]);}t = 0;for (int i = 1; i <= n; i++)p[i] = Point(i, sumv[i]);for (int i = n-1; i > 0; i--) {while (t > 1 && getY(Q[t], i+1)*getX(Q[t-1], Q[t]) >= getY(Q[t-1], Q[t])*getX(Q[t], i+1))t--;Q[++t] = i+1;int l = 1, r = t;while (l < r) {int mid = (l+r) >> 1;if (check2(mid, i))  r = mid;else l = mid + 1; } ans = max(ans, tmp+(LL)Q[r]*a[i]-(LL)i*a[i]-sumv[Q[r]]+sumv[i]);}printf("%I64d", ans);    return 0;}

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