仓库建设

http://www.lydsy.com/JudgeOnline/problem.php?id=1096

这道题和那道锯木厂选址很像,

按照自己的思维习惯我把所有的点翻转了，即从山底到山顶.

dist[i]表示第i个点到第一个点的距离,c[i]为在第i个点建的费用,p[i]表示第i个点的存货 

dp[i] = min(dp[j-1]+∑p[k]*(dist[k]-dist[j])+c[j])

设sp[i] = ∑p[k] (1 <= k <= i)

   spd[i] = ∑p[k]*dist[k] (1 <= k <= i)

  dp[j-1]+∑p[k]*(dist[k]-dist[j])+c[j]

= dp[j-1]+spd[i]-spd[j-1]-(sp[i]-sp[j-1])*dist[j]+c[j]

= dp[j-1]+spd[j-1]+sp[j-1]*dist[j]+c[j]-sp[i]*dist[j]+spd[i]

用y表示dp[j-1]+spd[j-1]+sp[j-1]*dist[j]+c[j]

x表示dist[j]

a表示sp[i]

就变成了斜率优化的形式G = -ax+y, a单调递增

dp[i] = min(G)+spd[i]

 
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>   
#include <map>
#include <string>  
#include <climits> 
#include <set>
#include <string>    
#include <sstream>
#include <utility>   
#include <ctime>
#include <bitset>
 
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::stringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
using std::unique;
using std::lower_bound;
using std::random_shuffle;
using std::bitset;
using std::upper_bound;
using std::multiset;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
typedef LL TY;
typedef long double LF;
 
const int MAXN(1000010);
const int MAXM(100010);
const int MAXE(100010);
const int MAXK(6);
const int HSIZE(31313);
const int SIGMA_SIZE(26);
const int MAXH(19);
const int INFI((INT_MAX-1) >> 1);
const ULL BASE(31);
const LL LIM(10000000);
const int INV(-10000);
const int MOD(20100403);
const double EPS(1e-7);
const LF PI(acos(-1.0));
 
template<typename T> void checkmax(T &a, T b){if(b > a) a = b;}
template<typename T> void checkmin(T &a, T b){if(b < a) a = b;}
template<typename T> T ABS(const T &a){return a < 0? -a: a;}
 
int que[MAXN];
int front, back;
LL dist[MAXN], spd[MAXN], sp[MAXN], table[MAXN];
int C[MAXN];
 
LL Y(int ind){ return table[ind-1]+C[ind]+sp[ind-1]*dist[ind]-spd[ind-1];}
LL X(int ind){ return dist[ind];}
 
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i = n; i >= 1; --i)
            scanf("%lld%lld%d", dist+i, sp+i, C+i);
        LL temp = dist[1];
        for(int i = 1; i <= n; ++i)
        {
            dist[i] = temp-dist[i];
            spd[i] = spd[i-1]+sp[i]*dist[i];
            sp[i] += sp[i-1];
        }
        int ti = 1;
        while(sp[ti] == 0) table[ti++] = 0;
        front = 0;
        back = -1;
        que[++back] = ti;
        for(int i = ti; i <= n; ++i)
        {
            LL ty = Y(i), tx = X(i);
            LL a = sp[i];
            while(back-front > 0 && (ty-Y(que[back-1]))*(tx-X(que[back])) >= (ty-Y(que[back]))*(tx-X(que[back-1]))) --back;
            que[++back] = i;
            while(back-front > 0 && (-a)*X(que[front+1])+Y(que[front+1]) <= (-a)*X(que[front])+Y(que[front])) ++front;
            table[i] = (-a)*X(que[front])+Y(que[front])+spd[i];
        }
        printf("%lld\n", table[n]);
    }
    return 0;
}