BZOJ 1010 [HNOI2008]玩具装箱toy (斜率优化DP)

1010: [HNOI2008]玩具装箱toy

Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 11699 Solved: 4956
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Description

　　P教授要去看奥运，但是他舍不下他的玩具，于是他决定把所有的玩具运到北京。他使用自己的压缩器进行压
缩，其可以将任意物品变成一堆，再放到一种特殊的一维容器中。P教授有编号为1...N的N件玩具，第i件玩具经过
压缩后变成一维长度为Ci.为了方便整理，P教授要求在一个一维容器中的玩具编号是连续的。同时如果一个一维容
器中有多个玩具，那么两件玩具之间要加入一个单位长度的填充物，形式地说如果将第i件玩具到第j个玩具放到一
个容器中，那么容器的长度将为 x=j-i+Sigma(Ck) i<=K<=j 制作容器的费用与容器的长度有关，根据教授研究，
如果容器长度为x,其制作费用为(X-L)^2.其中L是一个常量。P教授不关心容器的数目，他可以制作出任意长度的容
器，甚至超过L。但他希望费用最小.

Input

　　第一行输入两个整数N，L.接下来N行输入Ci.1<=N<=50000,1<=L,Ci<=10^7

Output

　　输出最小费用

Sample Input

5 4
3
4
2
1
4

Sample Output

HINT

析：很容易看出来是DP，而且方程也很好写，dp[i] = min{ dp[j] + (sum[i] - sum[j] + i - j - 1 - L)^2 }，然而复杂度是O(n^2)，肯定不行，但可以用斜率进行优化，先设f[i] = sum[i] + i，m = L + 1 ，那么这个方程都简单多了dp[j] + (f(i)-f(j)-m)^2，然后使用斜率优化就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 5e4 + 10;
const int maxm = 3e5 + 10;
const int mod = 100003;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn];
LL sum[maxn], dp[maxn];
int q[maxn];

LL f(int i){ return sum[i] + i; }

LL UP(int i, int j){ return dp[i]-dp[j] + sqr(f(i)) - sqr(f(j)); }
LL DOWN(int i, int j){ return 2 * (f(i) - f(j)); }
LL DP(int i, int j){ return dp[j] + sqr(f(i)-f(j)-m); }

int main(){
  scanf("%d %d", &n, &m);  ++m;
  for(int i = 1; i <= n; ++i){
    int x;  scanf("%d", &x);
    sum[i] = sum[i-1] + x;
  }
  dp[0] = 0;
  int fro = 0, rear = 0;
  q[++rear] = 0;
  for(int i = 1; i <= n; ++i){
    while(fro + 1 < rear && UP(q[fro+2], q[fro+1]) <= DOWN(q[fro+2], q[fro+1]) * (f(i) - m))  ++fro;
    dp[i] = DP(i, q[fro+1]);
    while(fro + 1 < rear && UP(i, q[rear]) * DOWN(i, q[rear-1]) < UP(i, q[rear-1]) * DOWN(i, q[rear])) --rear;
    q[++rear] = i;
  }
  printf("%lld
", dp[n]);
  return 0;
}