BZOJ 1096 ZJOI 2007 仓库建设 斜率优化DP

题目大意：山坡上有一些仓库，下雨了，要把所有仓库中的东西转移出来，每转移一单位的东西走一个单位长度需要花费1，在i处建立一个仓库需要花费cost[i]，求最小的花费。

思路：和小P的牧场好像啊。。。

记录两个前缀和，sum[i] = Σsrc[i]

_sum[i] = Σsrc[i] * pos[i],

然后DP方程：f[i] = f[j] + (sum[i] - sum[j]) * pos[i] - _sum[i] + _sum[j] + cost[i]

注意转换long long 数据有可能乘爆

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;
int pos[MAX],num[MAX],cost[MAX];
long long sum[MAX],_sum[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 >= 2)
        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 >= 2)
        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%d%d",&pos[i],&num[i],&cost[i]);
        sum[i] = sum[i - 1] + num[i];
        _sum[i] = _sum[i - 1] + (long long)pos[i] * num[i];
    }
    memset(f,0x3f,sizeof(f));
    f[0] = 0;
    for(int i = 1; i <= cnt; ++i) {
        Insert(sum[i - 1],f[i - 1] + _sum[i - 1]);
        Point ans = GetAns(pos[i]);
        f[i] = ans.y + sum[i] * pos[i] - ans.x * pos[i] - _sum[i] + cost[i];
    }
    cout << f[cnt] << endl;
    return 0;
}