BZOJ 3156 防御准备 斜率优化DP

标签：bzoj   斜率优化dp   动态规划   dp   

题目大意：给出一排东西，现在要建造防御塔，在i处建造防御塔的花费是cost[i]，所有东西的花费是他它距离右侧最近的防御塔的距离。求最小花费。

思路：很简单的斜率优化。DP方程：f[i] = f[j] + (i - j - 1) * (i - j) / 2 + cost[i]

然后简单整理一下会发现f[j] + (j + 1) * j / 2 = f[i] - i ^ 2 + i * j + i / 2 + cost[i]

之后移项，得到y = f[j] + j * (j + 1) / 2;

k = i,x = j;

然后斜率优化一下。。。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1000010
using namespace std;
 
struct Point{
    long long x,y;
     
    Point(long long _ = 0,long long __ = 0):x(_),y(__) {}
}q[MAX];
 
int cnt,cost[MAX];
long long f[MAX];
int front,tail;
 
inline double GetSlope(const Point &a,const Point &b)
{
    if(a.x == b.x)  return 1e15;
    return (double)(a.y - b.y) / (a.x - b.x);
}
 
inline void Insert(long long x,long long y)
{
    Point now(x,y);
    while(tail - front > 1)
        if(GetSlope(q[tail],now) < GetSlope(q[tail - 1],q[tail]))
            --tail;
        else    break;
    q[++tail] = now;
}
 
inline Point GetAns(double slope)
{
    while(tail - front > 1)
        if(GetSlope(q[front + 1],q[front + 2]) < slope)
            ++front;
        else    break;
    return q[front + 1];
}
 
int main()
{
    cin >> cnt;
    for(int i = 1; i <= cnt; ++i)
        scanf("%d",&cost[i]);
    for(int i = 1; i <= cnt; ++i) {
        Insert(i - 1,f[i - 1] + (long long)i * (i - 1) / 2);
        Point ans = GetAns(i);
        f[i] = ans.y + (long long)i * (i - 1) / 2 - (long long)i * ans.x + cost[i];
    }
    cout << f[cnt] << endl;
    return 0;
}

BZOJ 3156 防御准备 斜率优化DP

标签：bzoj   斜率优化dp   动态规划   dp   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41679615