hdu 6140 Hybrid Crystals

Hybrid Crystals

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 233    Accepted Submission(s): 134

Problem Description

> Kyber crystals, also called the living crystal or simply the kyber, and known as kaiburr crystals in ancient times, were rare, Force-attuned crystals that grew in nature and were found on scattered planets across the galaxy. They were used by the Jedi and the Sith in the construction of their lightsabers. As part of Jedi training, younglings were sent to the Crystal Caves of the ice planet of Ilum to mine crystals in order to construct their own lightsabers. The crystal's mix of unique lustre was called "the water of the kyber" by the Jedi. There were also larger, rarer crystals of great power and that, according to legends, were used at the heart of ancient superweapons by the Sith.
>
> — Wookieepedia

Powerful, the Kyber crystals are. Even more powerful, the Kyber crystals get combined together. Powered by the Kyber crystals, the main weapon of the Death Star is, having the firepower of thousands of Star Destroyers.

Combining Kyber crystals is not an easy task. The combination should have a specific level of energy to be stablized. Your task is to develop a Droid program to combine Kyber crystals.

Each crystal has its level of energy (i-th crystal has an energy level of ai). Each crystal is attuned to a particular side of the force, either the Light or the Dark. Light crystals emit positive energies, while dark crystals emit negative energies. In particular,

* For a light-side crystal of energy level ai, it emits +ai units of energy.
* For a dark-side crystal of energy level ai, it emits −ai units of energy.

Surprisingly, there are rare neutral crystals that can be tuned to either dark or light side. Once used, it emits either+ai or −ai units of energy, depending on which side it has been tuned to.

Given n crystals' energy levels ai and types bi (1≤i≤n),bi=N means the i-th crystal is a neutral one, bi=L means a Light one, and bi=D means a Dark one. The Jedi Council asked you to choose some crystals to form a larger hybrid crystal. To make sure it is stable, the final energy level (the sum of the energy emission of all chosen crystals) of the hybrid crystal must be exactlyk.

Considering the NP-Hardness of this problem, the Jedi Council puts some additional constraints to the array such that the problem is greatly simplified.

First, the Council puts a special crystal of a1=1,b1=N.

Second, the Council has arranged the other n−1 crystals in a way that

ai≤∑j=1i−1aj[bj=N]+∑j=1i−1aj[bi=L∩bj=L]+∑j=1i−1aj[bi=D∩bj=D](2≤i≤n).

[cond] evaluates to 1 if cond holds, otherwise it evaluates to 0.

For those who do not have the patience to read the problem statements, the problem asks you to find whether there exists a setS⊆{1,2,…,n} and values si for all i∈S such that

∑i∈Sai∗si=k,

where si=1 if the i-th crystal is a Light one, si=−1 if the i-th crystal is a Dark one, and si∈{−1,1} if the i-th crystal is a neutral one.

 

Input

The first line of the input contains an integer T, denoting the number of test cases.

For each test case, the first line contains two integers n (1≤n≤103) and k (|k|≤106).

The next line contains n integer a1,a2,...,an (0≤ai≤103).

The next line contains n character b1,b2,...,bn (bi∈{L,D,N}).

 

Output

If there exists such a subset, output "yes", otherwise output "no".

 

Sample Input

25 9 1 1 2 3 4N N N N N 6 -101 0 1 2 3 1N L L L L D

 

Sample Output

yesno

 

题意:

 'N'表示a[i] 可正可负（看需要），‘L’表示a[i]是正的，‘D’表示a[i]是负的，问在a[1...n]中任意取几个数相加使和为k，若能输出yes,否则no

序列a[]的条件是a[1]=1(b[i]=N，可正可负)

之后的若b[i]==N,  a[i]<=sum(a[j]  (b[j]==N))，j∈[1...i-1]

b[i]==D,a[i]<=sum(a[j]  (b[j]==N || b[j]==D)) ,j∈[1..i-1]

b[i]==L,a[i]<=sum( a[j]   (b[j]==N  || b[j]==L ) )  j∈[1..i-1]

解析：

一道大大大大大大大大大的水题。。。。。。。。

最后一段真的坑人。。。。。。。

只要把正的加+可正可负的lsum，负的+可正可负的dum

若k<0 将他与dsum比较，若dsum>=(-k)   则yes

若k>0将他与lsum比较，若lum>=k  则yes

证明主要因为他组成序列的条件使得，若k<=lsum,则序列一定可以在b[j]=L||b[j]==N的数中取出几个数组成k（因为序列是 <=1,<=2,<=4,<=8..）

#include<stdio.h>#include<string.h>int a[1010];int main(){int t,n,k,lsum,dsum;char c;scanf("%d",&t);while(t--){lsum=dsum=0;scanf("%d%d",&n,&k);for(int i=0;i<n;i++){scanf("%d",&a[i]);}getchar();for(int i=0;i<n;i++){scanf("%c%*c",&c);if(c=='L') lsum+=a[i];if(c=='D') dsum+=a[i];if(c=='N') lsum+=a[i],dsum+=a[i];}if(k>0){if(k<=lsum) printf("yes\n");else printf("no\n");}else{k=-k;if(k<=dsum) printf("yes\n");else printf("no\n");}}return 0;}