4 20 1

1 3 5

1 7 3

4 6 9

3 5 13

3

題解

• 最小值最大，顯然二分
• 現在二分出來一個最低高度，思考一下怎麼判斷
• 考慮dp，設f[i][s]為當前到第i位前k個位用泥土的狀態為s(是否有用)
• 那麼就很顯然了，每次轉移前現算出前k個給當前位的貢獻，設為num
• 轉移就有兩種一種是當前位不選，一種是當前位選
• 狀態轉移方程就很顯然了
• ①f[i+1][j>>1]=min(f[i+1][j>>1],f[i][j]);
  ②f[i+1][(j>>1)|(1<<(k-1))]=min(f[i+1][(j>>1)|(1<<(k-1))],f[i][j]+c[i+1]);
  

程式碼

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 int n,m,k,sum,mx,num,h[110],e[110],c[110],f[110][1<<12];
 6 bool check(int x)
 7 {
 8     memset(f,127,sizeof(f)),f[0][0]=0;
 9     for (int i=0;i<=n-1;i++)
10         for (int j=0;j<=(1<<k)-1;j++)
11         {
12             num=0;
13             for (int z=1;z<=k-1;z++) if (j&(1<<z)) num+=e[i-k+z+1];
14             if (num+h[i+1]>=x) f[i+1][j>>1]=min(f[i+1][j>>1],f[i][j]);
15             if (num+h[i+1]+e[i+1]>=x) f[i+1][(j>>1)|(1<<(k-1))]=min(f[i+1][(j>>1)|(1<<(k-1))],f[i][j]+c[i+1]);
16         }
17     for (int i=0;i<=(1<<k)-1;i++) if (f[n][i]<=m) return true;
18     return false;
19 }
20 int main()
21 {
22     freopen("cover.in","r",stdin),freopen("cover.out","w",stdout);
23     scanf("%d%d%d",&n,&m,&k);
24     for (int i=1;i<=n;i++) scanf("%d%d%d",&h[i],&e[i],&c[i]),sum+=e[i],mx=max(mx,h[i]);
25     int l=1,r=sum+mx;
26     while (l<r)
27     {
28         int mid=(l+r)>>1;
29         if (check(mid)) l=mid+1; else r=mid;
30     }
31     printf("%d",l-1);
32 }