【四边形不等式】POJ1160[IOI2000]-Post Office

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 const int MAXV=305;
 8 const int MAXP=35;
 9 const int INF=0x7fffffff;
10 int v,p;
11 int dis[MAXV],sum[MAXV],w[MAXV][MAXV];//w[i][j]表示在[i,j]间建立一个邮局的最小代价 
12 int s[MAXV][MAXP],dp[MAXV][MAXP];
13 
14 void init()
15 {
16     scanf("%d%d",&v,&p);
17     sum[0]=0;
18     for (int i=1;i<=v;i++) scanf("%d",&dis[i]);
19     sort(dis+1,dis+v+1);
20     for (int i=1;i<=v;i++) sum[i]=dis[i]+sum[i-1];
21     for (int i=1;i<=v;i++)
22     {
23         w[i][i]=0;
24         for (int j=i+1;j<=v;j++)
25         {
26             if ((i+j)%2==0) w[i][j]=sum[j]-sum[(i+j)/2]-sum[(i+j)/2-1]+sum[i-1];
27                 else w[i][j]=sum[j]-sum[(i+j)/2]-sum[(i+j)/2-1]+sum[i-1]-dis[(i+j)/2];
28         }
29     }
30 }
31 
32 void solve()
33 {
34     memset(dp,127,sizeof(dp));
35     for (int i=1;i<=v;i++) dp[i][1]=w[1][i];
36     for (int j=2;j<=p;j++)
37     {
38         s[v+1][j]=v-1;
39         for (int i=v;i>=j;i--)
40         {
41             for (int k=s[i][j-1];k<=s[i+1][j];k++)
42             {
43                 if (dp[i][j]>dp[k][j-1]+w[k+1][i])//一开始这里敲成了w[k+1][j] 
44                 {
45                     dp[i][j]=dp[k][j-1]+w[k+1][i];
46                     s[i][j]=k;
47                 }
48             }
49         } 
50     }
51     printf("%d",dp[v][p]); 
52 }
53 
54 int main()
55 {
56     init();
57     solve();    
58     return 0;    
59 }