input1:
4 6
2 1
6 3
7 2
5 6
input2:
2 7
4 6
3 5
input3:
4 15
5 5
5 12
6 10
2 1

output1:
3
output2:
0
output3:
4

題解

• 這題可以用dfs過，數據比較水，200+ms，非常優秀
• 記錄個前綴和，剪枝一下

代碼

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 using namespace std;
 5 int n;
 6 long long
 
 ans,k,h[50],g[50],p[50],f[50],q[50];
 7 void dfs(int d,int last,long long sum)
 8 {
 9     if (d>n)
10     {
11         if (sum>=k) ans++;
12         return;
13     }
14     if (h[d]>=h[last]&&sum+g[d]>=k) ans+=f[d];
15     else 
16         if (h[d]>=h[last]&&sum+q[d]>=k) dfs(d+1,d,sum+g[d]);
17     if (sum+p[d+1]>=k) dfs(d+1,last,sum);
18 }
19 int main()
20 {
21     freopen("san.in","r",stdin);
22     freopen("san.out","w",stdout);
23     scanf("%d%lld",&n,&k);
24     for (int i=1;i<=n;i++) scanf("%lld%lld",&h[i],&g[i]);
25     f[n]=1,q[n]=g[n],p[n]=g[n],p[n+1]=0;
26     for (int i=n-1;i>0;i--) 
27     {
28         f[i]=1,q[i]=0,p[i]=p[i+1]+g[i];
29         for(int j=i+1;j<=n;j++) 
30             if (h[j]>=h[i]) 
31             {
32                 f[i]+=f[j];
33                 if (q[j]>q[i]) q[i]=q[j];
34             }
35         q[i]+=g[i];
36     }
37     dfs(1,0,0);
38     printf("%lld",ans);
39     return 0;
40 }

[前綴和][dfs] Jzoj P5797 跳躍