題意: 給出m個村莊及其距離，給出n個郵局，要求怎麽建n個郵局使代價最小。

析：一般的狀態方程很容易寫出，dp[i][j] = min{dp[i-1][k] + w[k+1][j]}，表示前 j 個村莊用 k 個郵局距離最小，w可以先預處理出來O(n^2)，但是這個方程很明顯是O(n^3)，但是因為是POJ，應該能暴過去。。= =，正解應該是對DP進行優化，很容易看出來，w是滿足四邊形不等式的，也可以推出來 s 是單調的，可以進行優化。

代碼如下：

#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 = 300 + 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;
}

int dp[35][maxn], w[maxn][maxn], s[35][maxn];
int a[maxn];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    for(int i = 1; i <= n; ++i){
      w[i][i] = 0;
      for(int j = i+1; j <= n; ++j)
        w[i][j] = w[i][j-1] + a[j] - a[i+j>>1];
    }
    memset(dp, INF, sizeof dp);
    memset(s, 0, sizeof s);
    for(int i = 1; i <= n; ++i)  dp[1][i] = w[1][i];
    for(int i = 2; i <= m; ++i){
      s[i][n+1] = n;
      for(int j = n; j >= i; --j)
        for(int k = s[i-1][j]; k <= s[i][j+1]; ++k)
          if(dp[i][j] > dp[i-1][k] + w[k+1][j]){
            dp[i][j] = dp[i-1][k] + w[k+1][j];
            s[i][j] = k;
          }
    }
    printf("%d\n", dp[m][n]);
  }
  return 0;
}

POJ 1160 Post Office (四邊形不等式優化DP)