Poj 1160 Post Office

Poj 1160 Post Office

I don't know about optimization. I only know the most violent method. O (V ^ 3) is actually less than Ms.

When DP [I] [J] [k] is set, it indicates that J post offices are put in consideration of the first I town, and the town where the last post office is located is KBefore KThe minimum distance between all towns and the nearest post office (Tm made such a detour for itself ..)

So there is DP [I] [J] [I] = min {DP [I-1] [J-1] [k] + dis [k] [I]} Where k <I

DP [I] [J] [k] = DP [I-1] [J] [k]

In this example, DIS [k] [I] indicates the sum of the minimum values of all towns in the [K, I] range from post office K to post office I, that is:

Dis [I] [J] = sum {min (X [k]-X [I], X [J]-X [k])} I <= k <= J

Considering the boundary condition, for DP [I] [J] [K], since one of the first towns is not put, then K's getting any value is illegal (because I represent 2333333 of the status ..), therefore, for any I, consider the case where j = 1 on one side, that is, DP [I] [1] [k] = dis [0] [K], default X [0] =-INF

The final answer is: min {DP [v] [p] [k] + dis [k] [v]} p <= k <= V

The rest of the above logic is correct, and I submitted it happily. tm actually told me that the memory has exceeded 300 × 300 × 30. How can this problem be solved !!! TM memory only gives 10000, there is no way to compress the size of the first dimension to 2 (because the current I is only related to the I-1 ..)

79 Ms, O (V ^ 3), I smiled happily

 1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype>10 #include <cstring>11 #include <cstdlib>12 #include <iostream>13 #include <algorithm>14 using namespace std;15 #define INF 10000000016 #define inf (-((LL)1<<40))17 #define lson k<<1, L, mid18 #define rson k<<1|1, mid+1, R19 #define mem0(a) memset(a,0,sizeof(a))20 #define mem1(a) memset(a,-1,sizeof(a))21 #define mem(a, b) memset(a, b, sizeof(a))22 #define FOPENIN(IN) freopen(IN, "r", stdin)23 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)24 template<class T> T CMP_MIN ( T a, T b ) { return a < b;   }25 template<class T> T CMP_MAX ( T a, T b ) { return a > b;   }26 template<class T> T MAX ( T a, T b ) { return a > b ? a : b; }27 template<class T> T MIN ( T a, T b ) { return a < b ? a : b; }28 template<class T> T GCD ( T a, T b ) { return b ? GCD ( b, a % b ) : a; }29 template<class T> T LCM ( T a, T b ) { return a / GCD ( a, b ) * b;       }30 template<class T> T SWAP( T& a, T& b ) { T t = a; a = b;  b = t; }31 32 //typedef __int64 LL;33 typedef long long LL;34 const int MAXN = 255;35 const int MAXM = 110000;36 const double eps = 1e-12;37 38 int V, P;39 int x[310], DP[3][35][310];40 int dis[310][310], pre[310];41 42 void initDis()43 {44         mem0(dis);mem0(pre);45         for(int i=1;i<=V;i++)46         for(int j=i;j<=V;j++)47         for(int k=i;k<=j;k++)48         dis[i][j] += MIN(x[k]-x[i], x[j]-x[k]);49         for(int i=1;i<=V;i++)for(int j=i;j>=1;j--)50                 pre[i] += x[i] - x[j];51 }52 53 int main()54 {55         //FOPENIN ( "in.txt" );56        //FOPENOUT("out.txt");57        while(~scanf("%d %d", &V, &P))58        {59                 for(int i=1;i<=V;i++)60                         scanf("%d", &x[i]);61                 initDis();62                 for(int i=0;i<2;i++)63                 for(int j=0;j<=MIN(i,P);j++)64                 for(int k=0;k<=i;k++){65                         DP[i][j][k] = INF;66                 }67                 int now = 0;68                 for(int i=1;i<=V;i++)69                 {70                         now = !now;71                         int s = 0;72                         for(int j=1;j<=i;j++)   DP[now][1][j] = pre[j];73                         for(int j=2;j<=MIN(i, P); j++)74                         {75                                 DP[now][j][i] = INF;76                                 for(int k=j-1;k<i;k++)if(DP[!now][j-1][k] != INF)77                                 {78                                         DP[now][j][i] = MIN(DP[now][j][i], DP[now][j-1][k] + dis[k][i]);79                                 }80                                 for(int k=j;k<i;k++) DP[now][j][k] = DP[!now][j][k];81                         }82                 }83                 int ans = INF;84                 for(int k=P;k<=V;k++)if(DP[now][P][k] != INF)85                 {86                         int s = 0;87                         for(int j = k + 1; j <= V; j ++ ) s += x[j] - x[k];88                         ans = MIN(ans, DP[now][P][k] + s);89                 }90                 printf("%d\n", ans);91        }92         return 0;93 }