E - Swap

E - Swap Editorial

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 500500 points

Problem Statement

Given is a string SS consisting of K, E, Y.

How many strings are there that can be obtained with at most KK swaps of two adjacent characters in SS?

Constraints

• 2 \leq |S| \leq 302≤∣S∣≤30
• 0 \leq K \leq 10^90≤K≤109
• SS consists of K, E, Y.

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Input

Input is given from Standard Input in the following format:

SS
KK

Output

Print the answer.

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Sample Input 1 Copy

Copy

KEY
1

Sample Output 1 Copy

Copy

3

With at most one swap, there are three strings that can be obtained: KEY, EKY, KYE.

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Sample Input 2 Copy

Copy

KKEE
2

Sample Output 2 Copy

Copy

4

With at most two swaps, there are four strings that can be obtained: KKEE, KEKE, EKKE, KEEK.

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Sample Input 3 Copy

Copy

KKEEYY
1000000000

Sample Output 3 Copy

Copy

90

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此类题也只能说很难去证明为什么这样做不会缺失或者重复

EYEK

一维代表位置，二维代表e的个数，三维代表y的个数，四维代表交换次数

而按照正确方法我们得到的是如下结果

dp[1][1][0][0]=1

dp[1][1][0][1]=0

dp[1][1][0][2]=0

dp[1][1][0][3]=0

也就是交换两次结果也是0，可是如果我们把第二个e交换到第一个位置也是一种方法，而本题正确做法确实找到离当前e最近的那一个e进行转移。这就很让人不解

EYE为例

dp[2][2][0][0]为 1，显然不正确

但其转移到 dp[3][2][1][1]的时候，Y遍历E，会遇见一个比它大的E，也就把次数给加了上去。所以我们这样做是正确的。在累积多个相同字母时，如果我们没遇见其它字母，那么答案就一直是1种，（即先不计算与不同字母的交换），直到我们把不同字母提到前面，才产生了步数的变化和方案的累加。而且这么多不合理的1并不会影响答案，因为交换次数是0，我们最终获取答案的时候，仅仅取dp[n]的。

# include 
# include
# include
# include
# include
# include
# include
# include
# include
# include
 
 
using namespace std;
typedef long long int ll;
 
const int N=32;
const int C=N*(N-1)/2;
 
string s;
int K;
 
ll dp[N][N][N][C];
 
vector<int>kp,ep,yp;
 
int main()
{
    cin>>s>>K;
 
    s=" "+s;
 
 
    int n=s.size()-1;
 
    K=min(K,n*(n-1)/2);
 
    for(int i=1; i<=n; i++)
    {
        if(s[i]=='K')
        {
            kp.push_back(i);
 
        }
        else if(s[i]=='E')
        {
            ep.push_back(i);
 
        }
        else if(s[i]=='Y')
        {
 
            yp.push_back(i);
 
        }
    }
 
    int ck=kp.size();
 
    int ce=ep.size();
 
    int cy=yp.size();
 
 
 
    dp[0][0][0][0]=1;
 
 
    for(int i=0; i
    {
        for(int e=0; e<=ce; e++)
        {
            for(int y=0; y<=cy; y++)
            {
                int kk=i-e-y;
 
 
                if(kk<0||kk>ck)
                    continue;
 
 
 
                for(int st=0; st<=K; st++)
                {
                    if(!dp[i][e][y][st])
                        continue;
 
 
 
                    if (kk < ck)
                    {
                        int moveSteps = 0; // 把i+1位置的K往左交换
                        for (int l = 0; l < e; l++) if (ep[l] > kp[kk]) moveSteps++;
                        for (int l = 0; l < y; l++) if (yp[l] > kp[kk]) moveSteps++;
 
                        if (st + moveSteps <= K) dp[i + 1][e][y][st + moveSteps] += dp[i][e][y][st];
                    }
 
                    if (e < ce)
                    {
                        int moveSteps = 0;
                        for (int l = 0; l < kk; l++) if (kp[l] > ep[e]) moveSteps++;
                        for (int l = 0; l < y; l++) if (yp[l] > ep[e]) moveSteps++;
 
                        if (st + moveSteps <= K) dp[i + 1][e + 1][y][st+ moveSteps] += dp[i][e][y][st];
                    }
 
                    if (y < cy)
                    {
                        int moveSteps = 0;
                        for (int l = 0; l < kk; l++) if (kp[l] > yp[y]) moveSteps++;
                        for (int l = 0; l < e; l++) if (ep[l] > yp[y]) moveSteps++;
 
                        if (st+ moveSteps <= K) dp[i + 1][e][y + 1][st+ moveSteps] += dp[i][e][y][st];
                    }
 
                }
            }
        }
    }
 
    //  EKEY
 
  //  cout<
 
 
    ll res = 0;
 
    for (int i = 0; i <= K; i++) res+= dp[n][ce][cy][i];
 
    cout << res << endl;
 
 
 
    return 0;
}