POJ 3709 K-Anonymous Sequence (斜率优化DP)

题意：有一个不递减的序列，现在要把这些数分成若干个部分，每部分不能少于m个数。每部分的权值为所有数减去该部分最小的数的和。求最小的总权值。

析：状态方程很容易写出来，dp[i] = min{dp[j] + sum[i] - sum[j] - (i-j)*a[j+1] }，然而这个复杂度是 O(n^2)的肯定要TLE，

用斜率进行优化，维护一个下凸曲线，注意这个题是有个限制就是至少有要m个是连续的，所以开始的位置是2*m，想想为什么。

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>
#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 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 = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 500000 + 10;
const int mod = 1e9 + 7;
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;
}

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

LL getUP(int i, int j){
  return dp[i] - sum[i] + (LL)i*a[i+1] - (dp[j] - sum[j] + (LL)j*a[j+1]);
}

LL getDOWN(int i, int j){
  return a[i+1] - a[j+1];
}

LL getDP(int i, int j){
  return dp[j] + sum[i] - sum[j] - (LL)(i-j)*a[j+1];
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i){
      scanf("%d", a+i);
      sum[i] = sum[i-1] + a[i];
    }
    int fro = 0, rear = 0;
    q[++rear] = m;
    for(int i = m; i < 2*m; ++i)  dp[i] = sum[i] - i * a[1];
    for(int i = m * 2; i <= n; ++i){  // notice
      while(fro+1 < rear && getUP(q[fro+2], q[fro+1]) <= i*getDOWN(q[fro+2], q[fro+1]))  ++fro;
      dp[i] = getDP(i, q[fro+1]);
      int k = i - m + 1;
      while(fro+1 < rear && getUP(k, q[rear])*getDOWN(k, q[rear-1]) <= getUP(k, q[rear-1])*getDOWN(k, q[rear]))  --rear;
      q[++rear] = k;
    }
    printf("%I64d
", dp[n]);
  }
  return 0;
}