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
解法：状压dp
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int n,m,h[100100],w[100100],v[100100],f[(1<<20)+10],g[(1<<20)+10]; 7 //h,w,v分别为牛的身高，体重以及最大的承受重量； 8 //f表示在当前状态下的最大的负重，g表示在当前状态下的重量； 9 int b[30],ans=-1000; 10 //b表示1<<(i-1)；ans表示最大值； 11 int main() 12 {b[1]=1; 13 for(int i=2;i<=20;i++) b[i]=b[i-1]<<1;//预处理 14 scanf("%d%d",&n,&m); 15 for(int i=1;i<=n;i++) scanf("%d%d%d",&h[i],&w[i],&v[i]); 16 int maxx=(1<<n)-1; 17 for(int i=0;i<=maxx;i++) f[i]=-100000; 18 f[0]=0x7fffffff; 19 memset(g,0,sizeof(g)); 20 /*下面的动规是从前一个状态向后一个状态推的， 21 由样例可知：我们几乎无法在当前状态下运用前一个的状态的值， 22 因为根本无法用方程计算出何时负重量会更新（并不是只依赖于最下面的牛的负重， 23 如果中间的牛承受不住上面的牛的重量，那么依然不可能成立） 24 所以我们只能固定当前的状态，枚举接下来的可以放的牛的状态。*/ 25 for(int x=0;x<=maxx;x++) 26 for(int i=1;i<=n;i++) 27 { 28 if((x&b[i]))continue; 29 int t=x|b[i]; 30 if(f[x]<w[i])continue; 31 int tt=min(f[x]-w[i],v[i]);//如果最下面的是负载重量100，向上放一个负载重为1的牛，那么就不能选择前面一个决策，否则会出现状态上的错误。 32 f[t]=max(f[t],tt); //求出f[t]; 33 g[t]=g[x]+h[i];//求出高度，减少运算量 34 if(g[t]>=m)ans=max(ans,f[t]); 35 } 36 if(ans<0)printf("Mark is too tall"); 37 else 38 printf("%d",ans); 39 return 0; 40 }
仔细想想，许多方程基本上都能反过来写；如P3118即可。只要勤于思考，应该就能想出来一种吧。。。