2017ACM暑期多校联合训练 - Team 8 1008 HDU 6140 Hybrid Crystals （模拟）

题目链接

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 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 si for all i∈S such that

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

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

题意：

有n个宝石，每个宝石有自身的能量值，但是能量值可能为正也可能为负，有一个代表能量值正负的标记：

N：该宝石上的能量可以为正也可以为负

L：该宝石上的能量为正

D：该宝石上的能量为负

问这所有的宝石能不能构成能量为k的一个值。

分析：

这题目真的是又臭又长，说一堆没用的废话，瞬间感觉自己又经历了一场六级的阅读理解。。。心累啊

只怪自己比赛的时候脑子不够用，竟然用深搜在写，不超才怪呢。

这道题中的数组所组成的数构成了一个连续的区间。

如果之前的一堆数能够构成 [−a,b]中所有的整数的话，这时候来了一个数x，如果x只能取正值的话，并且有x<=b，那么就能构成[−a,b+x]内的所有的整数。

如果x只能取负值的话，并且有x<=a，那么就能构成[−a-x,b]内的所有的整数。

如果x可正可负的话，并且有x<=min(a,b)，那么就能构成[−a-x,b+x]内的所有的整数。

有疑问的一点就是不是需要构成一个连续的区间吗？

这一点题目上的那个臭长的式子保证了这一点。

代码：

#include<iostream>
#include<stdio.h>
using namespace std;
int a[1009];
int main()
{
    int T,n,k,sum1,sum2;///sum1表示正数的和，sum2表示负数的和
    char ch;
    scanf("%d",&T);
    while(T--)
    {
        sum1=sum2=0;
        scanf("%d%d",&n,&k);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(int i=1; i<=n; i++)
        {
            scanf(" %c",&ch);
            if(ch=='N')
            {
                sum1+=a[i];
                sum2-=a[i];
            }
            else if(ch=='L')
                sum1+=a[i];
            else
                sum2-=a[i];
        }
        if(k>=sum2&&k<=sum1)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}