2017 HDU 6140 多校联合赛 Hybrid Crystals

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 aiai, it emits +ai+ai units of energy.
• For a dark-side crystal of energy level aiai, it emits −ai−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 nn crystals’ energy levels aiai and types bibi (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 exactly k.

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

[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 set S⊆{1,2,…,n} and values sisi for all i∈Ssuch that

where si=1 if the i-th crystal is a Light one, si=−1 if the ii-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 TT, denoting the number of test cases.

For each test case, the first line contains two integers n (1≤n≤1000) and k (|k|≤1000000).

The next line contains n integer a1,a2,…,an(0≤ai≤1000).

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

2

5 9
1 1 2 3 4
N N N N N

6 -10
1 0 1 2 3 1
N L L L L D

Sample Output

yes
no

开篇先吐槽一下，你们看到这~么长的题目，还有做下去的心情吗，出个题目，还得先听个故事，做个阅读理解= =，真的难受。

**题意：**t组数据，给出n和k，一共有n个水晶，水晶分为光明、黑暗、中立。下面给出每个水晶的能量，

然后再给出水晶的属性，L表示光明水晶，带有正能量；D代表黑暗水晶，带有负能量；N代表中立水晶，可

以变为光明水晶也可以变为黑暗水晶。让你挑几个水晶，是否能使水晶能量总和恰好达到k。yes/no。

重点：为了降低题目难度，所以有给出了限制条件：

1.第一个水晶一定是能量为1的中立水晶。

2.第i个光明水晶的能量一定小于前面所能达到的能量之和。第i个黑暗水晶一定大于前面所能达到的能量之

和。

题解：

初始区间为[-1，1]，然后把符合条件的水晶加进来，扩展区间边界，最后只要k在区间内，便能凑出。详解

请看大佬博客。

代码：

#include<cstdio>#include<cstring>using namespace std;int a[2000];char b[2000];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,k;        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",&b[i]);            getchar();        }        int left,right;        left=-1;        right=1;        for(int i=1; i<n; i++)        {            if(b[i]=='L')            {                if(a[i]<=right)                    right+=a[i];            }            else if(b[i]=='D')            {                if(-a[i]>=left)                    left-=a[i];            }            else            {                if(a[i]<=right)                    right+=a[i];                if(-a[i]>=left)                    left-=a[i];            }        }        if(k<=right&&k>=left)            printf("yes\n");        else            printf("no\n");    }    return 0;}