There are n candy boxes in front of Tania. The boxes are arranged in a row from left to right, numbered from 1 to n. The i-th box contains ri candies, candies have the color ci (the color can take one of three values — red, green, or blue). All candies inside a single box have the same color (and it is equal to ci).
Initially, Tanya is next to the box number s. Tanya can move to the neighbor box (that is, with a number that differs by one) or eat candies in the current box. Tanya eats candies instantly, but the movement takes one second.
If Tanya eats candies from the box, then the box itself remains in place, but there is no more candies in it. In other words, Tanya always eats all the candies from the box and candies in the boxes are not refilled.
It is known that Tanya cannot eat candies of the same color one after another (that is, the colors of candies in two consecutive boxes from which she eats candies are always different). In addition, Tanya’s appetite is constantly growing, so in each next box from which she eats candies, there should be strictly more candies than in the previous one.
Note that for the first box from which Tanya will eat candies, there are no restrictions on the color and number of candies.
Tanya wants to eat at least k candies. What is the minimum number of seconds she will need? Remember that she eats candies instantly, and time is spent only on movements.
Input
The first line contains three integers n, s and k (1≤n≤50, 1≤s≤n, 1≤k≤2000) — number of the boxes, initial position of Tanya and lower bound on number of candies to eat. The following line contains n integers ri (1≤ri≤50) — numbers of candies in the boxes. The third line contains sequence of n letters ‘R’, ‘G’ and ‘B’, meaning the colors of candies in the correspondent boxes (‘R’ for red, ‘G’ for green, ‘B’ for blue). Recall that each box contains candies of only one color. The third line contains no spaces.
Output
Print minimal number of seconds to eat at least k candies. If solution doesn’t exist, print “-1”.
Examples
inputCopy
5 3 10
1 2 3 4 5
RGBRR
outputCopy
4
inputCopy
2 1 15
5 6
RG
outputCopy
-1
Note
The sequence of actions of Tanya for the first example:
move from the box 3 to the box 2;
eat candies from the box 2;
move from the box 2 to the box 3;
eat candy from the box 3;
move from the box 3 to the box 4;
move from the box 4 to the box 5;
eat candies from the box 5.
Since Tanya eats candy instantly, the required time is four seconds.
题意：
有n个排成一排的糖，起始位置为s，吃掉一个位置的所有糖不需要时间，往左或右移一步需要1的时间，他只会吃比上次吃位置更多糖的位置的糖，问你最少花掉多少时间。举个例子，1 2 3 4 5 当他吃了1之后可以随便吃其他的地方，吃了4之后只能吃5或者不吃。
题解：
这个可以用最短路做，先把所有点都放到队列里，假设是从那个位置开始的，dp[i][j]表示在第i个位置吃了j个糖果需要走的最少路程。由于每次都吃比当前位置所有的糖的数量大的那个位置的糖，那么就不需要担心原来的位置会被重复走。
#include<bits/stdc++.h>
using namespace std;
#define pa pair<int,int>
#define mp(a,b) make_pair(a,b)
const int inf=1e9;
int v[55],dp[55][2505],inq[55][2505];
char col[55];
int main()
{
int n,s,k;
scanf("%d%d%d",&n,&s,&k);
for(int i=1;i<=n;i++)
scanf("%d",&v[i]);
scanf("%s",col+1);
queue<pa>Q;
for(int i=1;i<=n;i++)
for(int j=1;j<=2500;j++)
dp[i][j]=inf;
for(int i=1;i<=n;i++)
{
int dis=abs(i-s);
dp[i][v[i]]=dis;
Q.push(mp(i,v[i]));
inq[i][v[i]]=1;
}
while(!Q.empty())
{
int pos=Q.front().first,val=Q.front().second;
Q.pop();
for(int i=1;i<=n;i++)
{
if(v[i]>v[pos]&&col[i]!=col[pos])
{
if(dp[i][val+v[i]]>dp[pos][val]+abs(i-pos))
{
dp[i][val+v[i]]=dp[pos][val]+abs(i-pos);
if(inq[i][val+v[i]])
continue;
Q.push(mp(i,val+v[i]));
}
}
}
}
int ans=inf;
for(int i=1;i<=n;i++)
{
for(int j=k;j<=2500;j++)
ans=min(ans,dp[i][j]);
}
printf("%d\n",ans==inf?-1:ans);
return 0;
}