【CodeForces 118D】【DP】 Caesar's Legions 【n1种步兵n2种马兵，询问多少种排列方式使不超过k1种步兵相连k2种马兵相连】

最新推荐文章于 2019-10-25 16:50:42 发布

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本篇介绍了一个关于步兵与马兵排列的问题，目的是寻找所有合法的排列组合，使得连续出现的同类兵种数量不超过限定值。通过动态规划的方法解决了这一问题，并给出了具体的实现代码。

传送门：D. Caesar's Legions

描述：

Gaius Julius Caesar, a famous general, loved to line up his soldiers. Overall the army had n1 footmen and n2 horsemen. Caesar thought that an arrangement is not beautiful if somewhere in the line there are strictly more that k1 footmen standing successively one after another, or there are strictly more than k2 horsemen standing successively one after another. Find the number of beautiful arrangements of the soldiers.

Note that all n1 + n2 warriors should be present at each arrangement. All footmen are considered indistinguishable among themselves. Similarly, all horsemen are considered indistinguishable among themselves.

Input

The only line contains four space-separated integers n1, n2, k1, k2 (1 ≤ n1, n2 ≤ 100, 1 ≤ k1, k2 ≤ 10) which represent how many footmen and horsemen there are and the largest acceptable number of footmen and horsemen standing in succession, correspondingly.

Output

Print the number of beautiful arrangements of the army modulo 100000000 (108). That is, print the number of such ways to line up the soldiers, that no more than k1 footmen stand successively, and no more than k2 horsemen stand successively.

   Examples
 
     input
   
2 1 1 10

     output
   
1

     input
   
2 3 1 2

     output
   
5

     input
   
2 4 1 1

     output
   
0

Note

Let's mark a footman as 1, and a horseman as 2.

In the first sample the only beautiful line-up is: 121

In the second sample 5 beautiful line-ups exist: 12122, 12212, 21212, 21221, 22121

题意：

n1种步兵n2种马兵，询问多少种排列方式使不超过k1种步兵相连k2种马兵相连

思路：

dp[i][j]表示前面i个步兵j个马兵排列种数，再枚举一下本次放的步兵和马兵的个数就可以dp了，0表示步兵，1表示马兵。

技巧在于：不是一个一个放兵，而是轮流种类放兵

代码：

#include <bits/stdc++.h>
using  namespace  std;
#define mod 100000000
#define rep(i,k,n) for(int i=k;i<=n;i++)

int dp[105][105][2];

int  main(){
  int n1, n2, k1, k2;
  cin>>n1>>n2>>k1>>k2;
  dp[0][0][0] = dp[0][0][1] = 1;
  rep(i, 0, n1){
    rep(j, 0, n2){
      for(int k=1; k<=i && k<=k1; k++)
        dp[i][j][1] = (dp[i][j][1] + dp[i-k][j][0]) % mod;
      for(int k=1; k<=j && k<=k2; k++)
        dp[i][j][0] = (dp[i][j][0] + dp[i][j-k][1]) % mod;
    }
  } 
  cout<<(dp[n1][n2][0] + dp[n1][n2][1]) % mod<<endl;
  return 0;
}