题目描述
Farmer John and his herd are playing frisbee. Bessie throws the
frisbee down the field, but it's going straight to Mark the field hand
on the other team! Mark has height H (1 <= H <= 1,000,000,000), but
there are N cows on Bessie's team gathered around Mark (2 <= N <= 20).
They can only catch the frisbee if they can stack up to be at least as
high as Mark. Each of the N cows has a height, weight, and strength.
A cow's strength indicates the maximum amount of total weight of the
cows that can be stacked above her.
Given these constraints, Bessie wants to know if it is possible for
her team to build a tall enough stack to catch the frisbee, and if so,
what is the maximum safety factor of such a stack. The safety factor
of a stack is the amount of weight that can be added to the top of the
stack without exceeding any cow's strength.
FJ将飞盘抛向身高为H(1 <= H <= 1,000,000,000)的Mark，但是Mark
被N(2 <= N <= 20)头牛包围。牛们可以叠成一个牛塔，如果叠好后的高度大于或者等于Mark的高度，那牛们将抢到飞盘。
每头牛都一个身高，体重和耐力值三个指标。耐力指的是一头牛最大能承受的叠在他上方的牛的重量和。请计算牛们是否能够抢到飞盘。若是可以，请计算牛塔的最大稳定强度，稳定强度是指，在每头牛的耐力都可以承受的前提下，还能够在牛塔最上方添加的最大重量。
输入输出格式
输入格式：
INPUT: (file guard.in)
The first line of input contains N and H.
The next N lines of input each describe a cow, giving its height,
weight, and strength. All are positive integers at most 1 billion.
输出格式：
OUTPUT: (file guard.out)
If Bessie's team can build a stack tall enough to catch the frisbee, please output the maximum achievable safety factor for such a stack.
Otherwise output "Mark is too tall" (without the quotes).
输入输出样例
4 10 9 4 1 3 3 5 5 5 10 4 4 5
2
动规 状压DP
看到数据范围就是状压DP了吧233
有那么一阵子我有DFS可以剪枝强行卡过去的错觉，然而果然是错觉。
f[i]记录的是当前状态（状压当前有哪些牛）的最大安全因子是多大
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<queue> 6 using namespace std; 7 const int INF=0x3f3f3f3f; 8 const int mxn=30; 9 int read(){ 10 int x=0,f=1;char ch=getchar(); 11 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();} 13 return x*f; 14 } 15 int f[1<<20]; 16 int g[1<<20]; 17 struct node{ 18 int h,w,s; 19 }a[mxn]; 20 int n,H; 21 int ans=-1; 22 int main(){ 23 int i,j; 24 n=read();H=read();int smm=0; 25 for(i=0;i<n;i++){ 26 a[i].h=read();a[i].w=read();a[i].s=read(); 27 smm+=a[i].h; 28 } 29 if(smm<H){printf("Mark is too tall ");return 0;} 30 memset(f,-1,sizeof f); 31 f[0]=INF; 32 int ed=(1<<n)-1; 33 for(i=0;i<=ed;i++){ 34 for(j=0;j<n;j++){ 35 if((i>>j)&1)continue; 36 int v=i^(1<<j); 37 if(f[i]<a[j].w)continue; 38 int t=min(f[i]-a[j].w,a[j].s); 39 f[v]=max(f[v],t); 40 g[v]=g[i]+a[j].h;//累计高度 41 if(v && g[v]>=H)ans=max(ans,f[v]); 42 } 43 } 44 if(ans==-1)printf("Mark is too tall "); 45 else printf("%d ",ans); 46 return 0; 47 }