poj1160Post Office（DP）

http://poj.org/problem?id=1160

算水过的吧 四重循环没优化 CZ说爆可过 就爆了

dp[i][j] = min(dp[i][j],dp[i-1][g]-s) 第i个点建在第j个村庄上 s 是这个点比上个点少的距离

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<cmath>

 7 using namespace std;

 8 #define N 1010

 9 #define INF 0xfffffff

10 int h[N],dp[40][N];

11 int main()

12 {

13     int i,j,k,n,m,g;

14     scanf("%d%d",&n,&m);

15     for(i = 1; i <= n; i++)

16     scanf("%d",&h[i]);

17     for(i = 1; i <= n ; i++)

18     {

19         int s = 0;

20         for(j = 1; j  <= n ; j++)

21         s+=abs(h[j]-h[i]);

22         dp[1][i] = s;

23     }

24     for(i = 2 ; i <= m ; i++)

25         for(j = i ; j <= n ; j++)

26         {

27             dp[i][j] = INF;

28             for(g = i-1; g < j ; g++)

29             {

30                 int s=0,ss=0;

31                 for(k = g+1 ; k < j; k++)

32                 {

33                     s+=min(h[k]-h[g],h[j]-h[k]);

34                     ss+=(h[k]-h[g]);

35                 }

36                 dp[i][j] = min(dp[i][j],dp[i-1][g]-ss-(h[j]-h[g])*(n-j+1)+s);

37             }

38         }

39     int minz=INF;

40     for(i = m; i <= n ; i++)

41     minz = min(minz,dp[m][i]);

42     printf("%d\n",minz);

43     return 0;

44 }

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